What volume of 0.540 mol L\(^{-1} \) hydrochloric acid will react completely with 1.34 g of sodium carbonate?
- 11.7 mL
- 23.4 mL
- 29.9 mL
- 46.8 mL
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What volume of 0.540 mol L\(^{-1} \) hydrochloric acid will react completely with 1.34 g of sodium carbonate?
\(D\)
\(\ce{2HCl + Na2CO3 \rightarrow 2NaCl +H2CO3}\)
\(\ce{n(Na2CO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{1.34}{105.99} = 0.0126\ \text{mol}}\)
\(\ce{n(HCl) = 0.0126 \times 2 = 0.0253\ \text{mol}} \)
\(\text{Vol (HCl)}\ = \dfrac{0.0253}{0.540} = 0.0468\ \text{L} = 46.8\ \text{mL}\)
\(\Rightarrow D\)
The ammonium ion content of mixtures can be determined by boiling the mixture with a known excess of sodium hydroxide. This converts the ammonium ions into gaseous ammonia, which is removed from the system.
\( \ce{NH4^{+}(aq) + OH^{-}(aq) \rightarrow NH3(g) + H2O(l)} \)
The excess sodium hydroxide can then be titrated with an acid solution of known concentration.
A fertiliser containing ammonium ions was analysed as follows.
\( Titration \) | \(Volume \ \ce{HCl} \ \text{(mL)} \) |
1 | 22.65 |
2 | 22.05 |
3 | 22.00 |
4 | 21.95 |
Calculate the mass of ammonium ions in the sample of fertiliser. (5 marks)
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\(0.4671\ \text{g} \)
\(\text{Average titre (HCl)}\ =\dfrac{22.05+22.00+21.95}{3}=22.00\ \text{mL} = 0.02200\ \text{L} \)
\(\ce{n(HCl)} = \ce{c \times V} = 0.02200 \times 0.1102 = 2.424 \times 10^{-3}\ \text{mol} \)
\(\ce{n(NaOH \text{excess})} = 2.424 \times 10^{-3}\ \text{mol} \)
\(\text{In the 250 mL flask:}\)
\(\ce{n(NaOH \text{excess})} = \dfrac{250.0}{20.00} \times 2.424 \times 10^{-3} = 3.031 \times 10^{-2}\ \text{mol} \)
\(\ce{n(NaOH \text{total}) = \ce{c \times V} = 0.0500 \times 1.124 = 5.620 \times 10^{-2}\ \text{mol}} \)
\(\ce{n(NaOH\ \text{reacting with}\ NH4+) = 5.620 \times 10^{-2}-3.031 \times 10^{-2} = 2.589 \times 10^{-2}\ \text{mol}} \)
\(\ce{n(NH4+) = 2.589 \times 10^{-2}\ \text{mol}} \)
\(\ce{MM(NH4+) = 14.01 + 4 \times 1.008 = 18.042} \)
\(\ce{m(NH4+) = 2.589 \times 10^{-2} \times 18.042 = 0.4671\ \text{g}} \)
The flowchart shown outlines the sequence of steps used to determine the concentration of an unknown hydrochloric acid solution.
Describe steps A, B and C including correct techniques, equipment and appropriate calculations. Determine the concentration of the hydrochloric acid. (8 marks)
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Step A
→ Prepare \(\ce{Na2CO3}\) by drying. Protect solid \(\ce{Na2CO3}\) from moisture in the air by storing in a desiccator.
→ Calculate mass of dried \(\ce{Na2CO3}\) required and weigh accurately.
\(\ce{m(Na2CO3) = 0.1 \times 0.5 \times 105.99 = 5.30 g}\)
→ Clean and rinse a 500 mL volumetric flask with distilled water.
→ Add 5.30 grams of \(\ce{Na2CO3}\) to the volumetric flask using a funnel and wash funnel using distilled water. Add distilled water to the flask to the bottom of the meniscus.
Step B
→ Clean and rinse a 50 mL burette. Fill burette with the unknown acid and place on a retort stand.
→ Clean and rinse a 250 mL conical flask with distilled water.
→ Clean a 25 mL pipette and rinse with 0.1 M \(\ce{Na2CO3}\) solution. Fill pipette with \(\ce{Na2CO3}\) solution to bottom of meniscus.
→ Transfer all pipette solution into conical flask and add an appropriate indicator. A white background (tile) should be placed under the flask to highlight any colour changes in the solution.
→ Slowly add acid solution from the burette into the conical flask and record the volume used when the indicator changes colour.
Step C
→ The initial titration represents a test run to establish an indicative volume. Three subsequent titrations should be performed with the average titration forming the basis of \(\ce{HCl}\) concentration calculations.
→ Calculate the concentration of \(\ce{HCl}\)
\(\ce{2H^+(aq) + CO3^{2-} (aq) \rightarrow H2CO3 (aq) \rightarrow H2O (l) + CO2 (g)}\)
\(\ce{M(Na2CO3) = 0.1 \times 0.025 = 2.5 \times 10^{-3}}\)
\(\ce{M(HCl) = 2 \times M(Na2CO3) = 5\times 10^{-3}}\)
\[\ce{[HCl] = \frac{M(HCl)}{Vol HCl} = \frac{5 \times 10^{-3}}{21.4 \times 10^{-3}} = 0.234 mol L^{-1}}\]
Step A
→ Prepare \(\ce{Na2CO3}\) by drying. Protect solid \(\ce{Na2CO3}\) from moisture in the air by storing in a desiccator.
→ Calculate mass of dried \(\ce{Na2CO3}\) required and weigh accurately.
\(\ce{m(Na2CO3) = 0.1 \times 0.5 \times 105.99 = 5.30 g}\)
→ Clean and rinse a 500 mL volumetric flask with distilled water.
→ Add 5.30 grams of \(\ce{Na2CO3}\) to the volumetric flask using a funnel and wash funnel using distilled water. Add distilled water to the flask to the bottom of the meniscus.
Step B
→ Clean and rinse a 50 mL burette. Fill burette with the unknown acid and place on a retort stand.
→ Clean and rinse a 250 mL conical flask with distilled water.
→ Clean a 25 mL pipette and rinse with 0.1 M \(\ce{Na2CO3}\) solution. Fill pipette with \(\ce{Na2CO3}\) solution to bottom of meniscus.
→ Transfer all pipette solution into conical flask and add an appropriate indicator. A white background (tile) should be placed under the flask to highlight any colour changes in the solution.
→ Slowly add acid solution from the burette into the conical flask and record the volume used when the indicator changes colour.
Step C
→ The initial titration represents a test run to establish an indicative volume. Three subsequent titrations should be performed with the average titration forming the basis of \(\ce{HCl}\) concentration calculations.
→ Calculate the concentration of \(\ce{HCl}\)
\(\ce{2H^+(aq) + CO3^{2-} (aq) \rightarrow H2CO3 (aq) \rightarrow H2O (l) + CO2 (g)}\)
\(\ce{M(Na2CO3) = 0.1 \times 0.025 = 2.5 \times 10^{-3}}\)
\(\ce{M(HCl) = 2 \times M(Na2CO3) = 5\times 10^{-3}}\)
\[\ce{[HCl] = \frac{M(HCl)}{Vol HCl} = \frac{5 \times 10^{-3}}{21.4 \times 10^{-3}} = 0.234 mol L^{-1}}\]
The concentration of hydrochloric acid in a solution was determined by an acid base titration using a standard solution of sodium carbonate.
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a. → \(\ce{Na2CO3}\) is a stable compound.
→ \(\ce{Na2CO3}\) is a pure solid that will not readily absorb water from the atmosphere.
→ An accurate weight of \(\ce{Na2CO3}\) can therefore be obtained in the experiment’s measurements.
b. 0.2425 mol L–¹
c. → This is a strong acid / weak base titration.
→ Its equivalence point will occur at a pH less than seven and phenolphthalein changes colour in the pH range 10 – 8.3.
→ Phenolphthalein indicator would therefore signal the end point before equivalence (i.e. with a lower volume of acid).
→ The calculated concentration of \(\ce{HCl}\) would be higher than the correct concentration.
a. → \(\ce{Na2CO3}\) is a stable compound.
→ \(\ce{Na2CO3}\) is a pure solid that will not readily absorb water from the atmosphere.
→ An accurate weight of \(\ce{Na2CO3}\) can therefore be obtained in the experiment’s measurements.
b. \(\ce{Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)}\)
\[\ce{Average titre = \frac{21.65 + 21.70 + 21.60}{3} = 21.65 mL}\]
\[\ce{n(Na2CO3) = c \times V = 0.1050 \times 0.0250 = 0.002625 mol}\]
\(\ce{n(HCl) = 2 \times n(Na2CO3) = 0.005250 mol}\)
\[\ce{[HCl] = \frac{n}{V} = \frac{0.005250}{0.02165} = 0.2425 mol L^{-1}}\]
c. → This is a strong acid / weak base titration.
→ Its equivalence point will occur at a pH less than seven and phenolphthalein changes colour in the pH range 10 – 8.3.
→ Phenolphthalein indicator would therefore signal the end point before equivalence (i.e. with a lower volume of acid).
→ The calculated concentration of \(\ce{HCl}\) would be higher than the correct concentration.
A solution of hydrochloric acid was standardised by titration against a sodium carbonate solution using the following procedure.
The titration was performed and the hydrochloric acid was found to be 0.200 mol L¯1.
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• A 0.145 g sample of the seashell was placed in a conical flask.
• 50.0 mL of the standardised hydrochloric acid was added to the conical flask.
• At the completion of the reaction, the mixture in the conical flask was titrated with 0.250 mol L¯1 sodium hydroxide.
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a. → Distilled water.
→ This should be used to rinse the conical flask as this will not change the number of moles of \(\ce{Na2CO3}\) placed in it.
b. 54.3%
a. → Distilled water.
→ This should be used to rinse the conical flask as this will not change the number of moles of \(\ce{Na2CO3}\) placed in it.
b. \(\ce{HCl(aq) + NaOH(aq) -> NaCl (aq) + H2O(l)}\)
\(\ce{n(NaOH) = c \times V = 0.250 \times 0.0295 = 7.375 \times 10^{-3} mol}\)
\(\ce{n(HCl)_{after} = n(NaOH) = 7.375 \times 10^{-3} mol}\)
\(\ce{n(HCl)_{orig} = c \times V = 0.200 \times 0.0500 = 0.010 mol}\)
\(\ce{(HCl)_{used} = 0.0100-7.375 \times 10^{-3} = 2.625 \times 10^{-3} mol}\)
\(\ce{2HCl(aq) + CO3^2-(aq) -> CO2(g) + 2Cl-(aq)}\)
\[\ce{n(CO3)^2- = \frac{1}{2} \times 2.625 \times 10^{-3} = 1.3125 \times 10^{-3} mol}\]
\(\ce{m(CO3^2-) = 1.3125 \times 10^{-3} \times (12.01 + 3 \times 16.00) = 0.07876 g}\)
\[\therefore \ce{\text{%}(CO3^2-) = \frac{0.07876}{0.145} \times 100 = 54.3\text{%}}\]
A sodium hydroxide solution was titrated against citric acid \(\ce{(C6H8O7)}\) which is triprotic.
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a.
b. Technology solution
→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.
→ The equivalence point would be identified by a steep rise in the pH on the graph.
c. \(\ce{C6H8O7 + 3NaOH -> C6H5O7Na3 + 3H2O}\)
\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 mol}\)
\(\ce{n(NaOH) = 3 \times 0.00250 = 0.00750 mol}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.181 mol L^{-1} (3 sig fig)}\]
b. Technology solution
→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.
→ The equivalence point would be identified by a steep rise in the pH on the graph.
c. \(\ce{C6H8O7 + 3NaOH -> C6H5O7Na3 + 3H2O}\)
\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 mol}\)
\(\ce{n(NaOH) = 3 \times 0.00250 = 0.00750 mol}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.181 mol L^{-1} (3 sig fig)}\]
The graph shows the changes in pH during a titration.
Which pH range should an indicator have to be used in this titration?
`D`
→ The indicator suitable for this titration needs to completely change colour at the centre of the vertical section of the graph (equivalence point).
→ The range that will achieve this is `8.3-10.0`.
`=>D`
A solution of sodium hydroxide was titrated against a standardised solution of acetic acid which had a concentration of 0.5020 mol L¯1.
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a. \(\ce{CH3COOH(aq) + NaOH(aq) -> NaCH3COO(aq) + H2O(l)}\)
b. \(\ce{CH3COO-(aq) + H2O \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)
→ The acetate ion is a weak base.
→ As a result, it has accepted a proton from the water resulting in production of hydroxide ions.
→ Therefore the solution has a pH > 7.
a. \(\ce{CH3COOH(aq) + NaOH(aq) -> NaCH3COO(aq) + H2O(l)}\)
\(\ce{n(CH3COOH) = c \times V = 0.5020 \times 0.0250 = 0.01255 mol}\)
\(\ce{n(NaOH) = n(CH3COOH) = 0.01255 mol}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.01255}{0.01930} = 0.6503 mol L^{-1}}\]
b. \(\ce{CH3COO-(aq) + H2O \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)
→ The acetate ion is a weak base.
→ As a result, it has accepted a proton from the water resulting in production of hydroxide ions.
→ Therefore the solution has a pH > 7.
Which type of glassware is used in a titration to deliver an accurate volume of a solution to a known volume of another solution?
`D`
A burette is used in a titration.
`=>D`
25.0 mL of a 0.100 mol L ¯1 acid is to be titrated against a sodium hydroxide solution until final equivalence is reached.
Which of the following acids, if used in the titration, would require the greatest volume of sodium hydroxide?
`B`
→ Citric acid is triprotic (i.e. ratio moles NaOH : acid = 3 : 1). It therefore requires the greates volume of NaOH.
→ Acetic acid and Hydrochloric acid are monoprotic (i.e. ratio moles NaOH : acid = 1 : 1)
→ Sulfuric acid is diprotic (i.e. ratio moles NaOH : acid = 2 : 1)
`=>B`
In an experiment, 30 mL of water is to be transferred into a conical flask.
Which piece of equipment would deliver the volume with the greatest accuracy?
`A`
→ A burette can measure and deliver to 0.05 mL accuracy, making it easily the most accurate measuring device of the given options.
`=>A`
A conductometric titration was undertaken to determine the concentration of a barium hydroxide solution. The solution was added to 250.0 mL of standardised 1.050 × 10 ¯ 3 mol L ¯1 hydrochloric acid solution. The results of the titration are shown in the conductivity graph.
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a. Titration curve shape:
→ The conductivity of a solution of hydrochloric acid is initially very high due to the high concentration of ions in the solution.
→ As barium hydroxide is added, the conductivity decreases because the concentration of \(\ce{H+}\) decreases due to the neutralization reaction with \(\ce{OH-}\) ions.
→ The conductivity reaches a minimum at the equivalence point, which is when all the \(\ce{H+}\) ions have been removed and the solution only contains \(\ce{Ba^2+}\) and \(\ce{Cl-}\) ions.
→ These ions are much less mobile than \(\ce{H+}\) or \(\ce{OH-}\) ions, so the conductivity is lower at the equivalence point.
→ After the equivalence point, the conductivity increases as more barium hydroxide is added, resulting in an increase in the concentration of \(\ce{OH-}\) ions and therefore an increase in conductivity.
b. \(\ce{2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2H2O(l)} \)
\[\ce{Ba(OH)2 }\] | `=7.653 xx 10^{-3}\ text{mol L}^{-1}` |
a. Titration curve shape:
→ The conductivity of a solution of hydrochloric acid is initially very high due to the high concentration of ions in the solution.
→ As barium hydroxide is added, the conductivity decreases because the concentration of \(\ce{H+}\) decreases due to the neutralization reaction with \(\ce{OH-}\) ions.
→ The conductivity reaches a minimum at the equivalence point, which is when all the \(\ce{H+}\) ions have been removed and the solution only contains \(\ce{Ba^2+}\) and \(\ce{Cl-}\) ions.
→ These ions are much less mobile than \(\ce{H+}\) or \(\ce{OH-}\) ions, so the conductivity is lower at the equivalence point.
→ After the equivalence point, the conductivity increases as more barium hydroxide is added, resulting in an increase in the concentration of \(\ce{OH-}\) ions and therefore an increase in conductivity.
b. \(\ce{2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2H2O(l)} \)
\(\ce{n(HCl) = c \times V = 1.050 \times 10^{-3} \times 250 = 2.625 \times 10^{-4} mol }\)
\(\ce{n(Ba(OH)2) = 0.5 \times n(HCl) = 1.3125 \times 10^{-4} mol}\)
\[\ce{[Ba(OH)2] = \frac{n}{V}}\] | `=(1.3125 xx 10^{-4})/(0.01715)` | |
`=7.653 xx 10^{-3}\ text{mol L}^{-1}` |
A chemist used the following method to determine the concentration of a dilute solution of propanoic acid `(pK_(a)=4.88)`.
The chemist weighed out 1.000 g of solid `text{NaOH}` on an electronic balance and then made up the solution in a 250.0 mL volumetric flask.
The chemist then performed titrations, using bromocresol green as the indicator. This indicator is yellow below pH 3.2 and green above pH 5.2.
The results are shown in the table.
Explain why this method produces inaccurate and unreliable results. (3 marks)
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→ `text{NaOH}` is hygroscopic and cannot be accurately used as a primary standard in titration experiments.
→ When the solid `text{NaOH}` is weighed, it will have absorbed water from the atmosphere, which means that the solution made from it will be more dilute than expected.
→ Furthermore, because this is a titration between a weak acid and a strong base which produces a basic salt, the pH at the equivalence point will be greater than 7.
→ Using Bromocresol green as an indicator in this case would not be suitable because it changes colour in the pH range of 3.2–5.2. This is the flat region of the titration curve, before the equivalence point.
→ The results of the titration are also unreliable because the indicator used produces a non-sharp endpoint, resulting in significantly different titres in each titration.
→ `text{NaOH}` is hygroscopic and cannot be accurately used as a primary standard in titration experiments.
→ When the solid `text{NaOH}` is weighed, it will have absorbed water from the atmosphere, which means that the solution made from it will be more dilute than expected.
→ Furthermore, because this is a titration between a weak acid and a strong base which produces a basic salt, the pH at the equivalence point will be greater than 7.
→ Using Bromocresol green as an indicator in this case would not be suitable because it changes colour in the pH range of 3.2–5.2. This is the flat region of the titration curve, before the equivalence point.
→ The results of the titration are also unreliable because the indicator used produces a non-sharp endpoint, resulting in significantly different titres in each titration.
The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined.
Step 1: A solution of \( \ce{NaOH(aq)} \) was standardised by titrating it against 25.00 mL aliquots of a solution of the monoprotic acid potassium hydrogen phthalate \( \ce{(KHP)} \). The \( \ce{(KHP)} \) solution was produced by dissolving 4.989 g in enough water to make 100.0 mL of solution. The molar mass of \( \ce{(KHP)} \) is 204.22 g mol ¯1.
The results of the standardisation titration are given in the table.
Step 2: A 75.00 mL bottle of the drink was opened and the contents quantitatively transferred to a beaker. The soft drink was gently heated to remove \( \ce{CO2}\).
Step 3: The cooled drink was quantitatively transferred to a 250.0 mL volumetric flask and distilled water was added up to the mark.
Step 4: 25.00 mL samples of the solution were titrated with the \( \ce{NaOH(aq)}\) solution. The average volume of \( \ce{NaOH(aq)} \) used was 13.10 mL.
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a. `0.1298 text{mol L}^(–1)`
b. \( \ce{CO2} \) can dissolve in water to produce \( \ce{H2CO3}\):
\( \ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq)} \)
→ This would enable \( \ce{NaOH} \) to react with \( \ce{H2CO3:} \)
\( \ce{2NaOH(aq) + H2CO3(aq) -> Na2CO3(aq) + 2H2O(l)} \)
→ Therefore, if \( \ce{CO2} \) was not removed, more \( \ce{NaOH} \) would be required to reach the endpoint.
→ This would result in a higher citric acid concentration calculation.
a. \( \ce{KHP(aq) + NaOH(aq) -> NaKP(aq) + H2O(l)} \)
`text{n(HX)}` | `= text{m} / text{MM}` | |
`= 4.989 / 204.22` | ||
`= 0.02443\ text{mol}` |
`[text(HX)]= n / V= 0.02443 / 0.1000= 0.2443 text{mol L}^(–1)`
`text{n(HX) titrated}` | `= text{c} xx text{V}` | |
`= 0.2443 xx 0.02500` | ||
`= 0.0006107\ text{mol}` |
`=>\ text{n(NaOH)}= 0.0006107 text{mol}`
Eliminate the first trial because it is an outlier.
`text{V}_(text(avg))text{(NaOH)}= 1 / 3 xx (27.40 + 27.20 + 27.60)= 27.40\ text{mL}`
`text{[NaOH]}= text{n} / text{V}= [6.107 xx 10^−3] / 0.02740 = 0.2229 text{mol L}^(–1)`
\( \ce{H3X(aq) + 3NaOH(aq) -> Na3X(aq) + 3H2O(l)} \)
`text{n(NaOH) titrated}` | `= text{c} xx text{V}` | |
`=0.2229 xx 0.01310` | ||
`= 2.920 xx 10^(−3)\ text{mol}` |
`text{n(H}_3 text{X)}= 1/3 xx 2.920 xx 10^(−3)= 9.733 xx 10^(−4)\ text{mol}`
`text{[H}_3 text{X] diluted} = text{n} / text{V} = (9.733 xx 10^(−4) )/ 0.025= 0.03893 \ text{mol L}^(–1)`
`text{[H}_3 text{X] 0riginal}` | `= 250.0 / 75.00 xx 0.03893` |
`= 0.1298 text{mol L}^(–1)` |
Therefore, the concentration of citric acid in the soft drink is 0.1298 mol L¯1.
b. \( \ce{CO2} \) can dissolve in water to produce \( \ce{H2CO3}\):
\( \ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq)} \)
This would enable \( \ce{NaOH} \) to react with \( \ce{H2CO3:} \)
\( \ce{2NaOH(aq) + H2CO3(aq) -> Na2CO3(aq) + 2H2O(l)} \)
→ Therefore, if \( \ce{CO2} \) was not removed, more \( \ce{NaOH} \) would be required to reach the endpoint.
→ This would result in a higher citric acid concentration calculation.
A manufacturer requires that its product contains at least 85% v/v ethanol.
The concentration of ethanol in water can be determined by a back titration. Ethanol is first oxidised to ethanoic acid using an excess of acidified potassium dichromate solution.
\(\ce{3C2H5OH($aq$) + 2Cr2O7^2-($aq$) + 16H^+($aq$) ->3CH3COOH($aq$) + 4Cr^3+($aq$) + 11H2O($l$)}\)
The remaining dichromate ions are reacted with excess iodide ions to produce iodine \(\ce{(I2)}\)
\(\ce{Cr2O7^2-($aq$) + 14H^+($aq$) + 61^-($aq$) -> 2Cr^3+($aq$) + 7H2O($l$) + 3I2($aq$)}\)
The iodine produced is then titrated with sodium thiosulfate \(\ce{(Na2S2O3)}\).
\(\ce{I2($aq$) + 2S2O3^2-($aq$) -> 2I^-($aq$) + S4O6^2-($aq$)}\)
A 25.0 mL sample of the manufacturer's product was diluted with distilled water to 1.00 L. A 25.0 mL aliquot of the diluted solution was added to 20.0 mL of 0.500 mol L¯1 acidified potassium dichromate solution in a conical flask. Potassium iodide (2.0 g)* was added and the solution titrated with 0.900 mol L¯1 sodium thiosulfate. This was repeated three times.
*The quantity of potassium iodide should be 5.0 g. This is not required for the calculation.
The following results were obtained.
The density of ethanol is 0.789 g mL¯1.
Does the sample meet the manufacturer's requirements? Support your answer with calculations. (7 marks)
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\( \text{V}_\text{avg}\ (\ce{Na2S2O3}) \) | `= (28.7 + 28.4 + 28.6) / 3` |
`= 28.5666… text{mL}` | |
`= 0.0285666… text{L}` |
→ The first titration is an outlier and so is excluded from the average.
`text{n(Na}_2 text{S}_2 text{O}_3text{)}=\ text{c} xx text{V}= 0.900 xx 0.0285666…= 0.02571 text{mol}`
\( \ce{n(S2O3^2-) = n(Na2S2O3) = 0.02571\ \text{mol}} \)
`text{I}_2 text{and S}_2 text{O}_3 ^(2-)\ text{are in a}\ 1 : 2\ text{ratio:}`
`text{n(I}_2text{)} = 1/2 xx text{n(S}_2 text{O}_3^( 2-) text{)}= 1/2 xx 0.02571= 0.012855\ text{mol}`
`text{Excess Cr}_2 text{O}_7^(\ \ 2-) text{and I}_2 text{are in}\ 1:3\ text{ratio:}`
`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}= 1/3 xx text{n(I}_2 text{)}= 1/3 xx 0.012855= 0.004285\ text{mol}`
`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}= text{c} xx text{V}= 0.500 xx 20/1000= 0.01\ text{mol}`
`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) reacted with ethanol}`
`= text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}-text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}`
`= 0.01-0.004285`
`= 0.005715\ text{mol}`
`text{n(C}_2 text{H}_5 text{OH)}= 3/2 xx text{n(Cr}_2 text{O}_7^(\ \ 2-) text{)}= 3/2 xx 0.005715= 0.0085725\ text{mol}`
`text{m(C}_2 text{H}_5 text{OH)}= text{n} xx text{MM}=0.0085725 xx (2 xx 12.01 + 6 xx 1.008 + 16.00)= 0.3949\ text{g}`
→ Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.
Find the mass of ethanol in the original solution:
`text{m(C}_2 text{H}_5 text{OH) original}= 0.3949… xx 1000/25= 15.796…\ text{g}`
`text{D}= text{m} / text{V}\ \ =>\ \ text{V}= text{m} / text{D}`
`text{V(C}_2 text{H}_5 text{OH)}= 15.796 / 0.789= 20.021\ text{mL}`
`text{% (C}_2text{H}_5text{OH)}= text{V(ethanol)} / text{V(sample)}= 20.021 / 25.0= 80.08… %\ text{v}//text{v}`
→ Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.
\( \text{V}_\text{avg}\ (\ce{Na2S2O3}) \) | `= (28.7 + 28.4 + 28.6) / 3` |
`= 28.5666… text{mL}` | |
`= 0.0285666… text{L}` |
→ The first titration is an outlier and so is excluded from the average.
`text{n(Na}_2 text{S}_2 text{O}_3text{)}=\ text{c} xx text{V}= 0.900 xx 0.0285666…= 0.02571 text{mol}`
\( \ce{n(S2O3^2-) = n(Na2S2O3) = 0.02571\ \text{mol}} \)
`text{I}_2 text{and S}_2 text{O}_3 ^(2-)\ text{are in a}\ 1 : 2\ text{ratio:}`
`text{n(I}_2text{)} = 1/2 xx text{n(S}_2 text{O}_3^( 2-) text{)}= 1/2 xx 0.02571= 0.012855\ text{mol}`
`text{Excess Cr}_2 text{O}_7^(\ \ 2-) text{and I}_2 text{are in}\ 1:3\ text{ratio:}`
`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}= 1/3 xx text{n(I}_2 text{)}= 1/3 xx 0.012855= 0.004285\ text{mol}`
`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}= text{c} xx text{V}= 0.500 xx 20/1000= 0.01\ text{mol}`
`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) reacted with ethanol}`
`= text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}-text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}`
`= 0.01-0.004285`
`= 0.005715\ text{mol}`
`text{n(C}_2 text{H}_5 text{OH)}= 3/2 xx text{n(Cr}_2 text{O}_7^(\ \ 2-) text{)}= 3/2 xx 0.005715= 0.0085725\ text{mol}`
`text{m(C}_2 text{H}_5 text{OH)}= text{n} xx text{MM}=0.0085725 xx (2 xx 12.01 + 6 xx 1.008 + 16.00)= 0.3949\ text{g}`
→ Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.
Find the mass of ethanol in the original solution:
`text{m(C}_2 text{H}_5 text{OH) original}= 0.3949… xx 1000/25= 15.796…\ text{g}`
`text{D}= text{m} / text{V}\ \ =>\ \ text{V}= text{m} / text{D}`
`text{V(C}_2 text{H}_5 text{OH)}= 15.796 / 0.789= 20.021\ text{mL}`
`text{% (C}_2text{H}_5text{OH)}= text{V(ethanol)} / text{V(sample)}= 20.021 / 25.0= 80.08… %\ text{v}//text{v}`
→ Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.
A student used the following method to titrate an acetic acid solution of unknown concentration with a standardised solution of dilute sodium hydroxide.
Compared to the actual concentration of the acetic acid, the calculated concentration will be
`B`
Two points to consider in method:
Burette rinsed with water instead of sodium hydroxide.
→ Sodium hydroxide solution diluted requiring more sodium hydroxide to neutralise the acetic acid.
This will result in greater moles of acetic acid, and thus the calculated concentration of acetic acid would be greater than the actual concentration.
Conical flask rinsed with acetic acid
→ Results in a greater number of moles of acid. More volume of sodium hydroxide would be added
→ Greater number of moles of acetic acid.
This would result in a greater number of moles, and thus the calculated concentration would be higher than the actual concentration.
In both cases, the calculated concentration would be higher than the actual concentration.
`=>B`