smc-3675-20-Titration Curves and Conductivity Graphs

CHEMISTRY, M6 2024 HSC 34

An aqueous solution of ammonia is added to a solution containing hydrochloric acid. A plot of conductivity against volume of ammonia solution added is shown. The temperature of the solution is kept constant throughout and the conductivity of the solution is corrected for dilution.

The relative conductivities of some relevant ions are shown in the table.

\begin{array}{|l|c|}
\hline \textit{Ion } & \textit{Relative conductivity } \\
\hline \ce{H^{+}} & 4.76 \\
\hline \ce{OH^{-}} & 2.70 \\
\hline \ce{Cl^{-}} & 1.04 \\
\hline \ce{NH_4^{+}} & 1.00 \\
\hline
\end{array}

Explain the shape of the graph. Include TWO balanced chemical equations in your answer. (4 marks)

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Initially, \(\ce{H^+(aq)}\) and \(\ce{Cl^-(aq)}\) are present in the solution. Due to the significant relative conductivity of \(\ce{H^+(aq)}\), the overall conductivity of the solution is high.
As ammonia is added to the solution prior to the equivalence point (below 4.5 mL), hydrochloric acid is neutralized by the addition of ammonia according to the reaction:
\(\ce{H^+(aq) + NH3(aq) -> NH4^+(aq)}\)
This results in the replacement of \(\ce{H^+}\), which exhibit high conductivity, with \(\ce{NH4^+}\), which have lower conductivity. Solution conductivity decreases.
The conductivity at the equivalence point is the lowest as only \(\ce{Cl^-}\) and \(\ce{NH4^+}\) ions are present which both have low relative conductivities.
Beyond the equivalence point, the excess ammonia added reacts partially with water to form \(\ce{NH4^+}\) and \(\ce{OH^-}\) ions according to the equation:
\(\ce{NH3(aq) + H2O(l) \rightleftharpoons NH4^+(aq) + OH^-(aq)}\)
Although these ions are more conductive than the reactant molecules, the ionisation of ammonia is limited due to the presence of \(\ce{NH4^+}\) ions already in the solution (as per Le Châtelier’s Principle).
Consequently, the conductivity increases only slightly after the equivalence point.

Show Worked Solution

Initially, \(\ce{H^+(aq)}\) and \(\ce{Cl^-(aq)}\) are present in the solution. Due to the significant relative conductivity of \(\ce{H^+(aq)}\), the overall conductivity of the solution is high.
As ammonia is added to the solution prior to the equivalence point (below 4.5 mL), hydrochloric acid is neutralized by the addition of ammonia according to the reaction:
\(\ce{H^+(aq) + NH3(aq) -> NH4^+(aq)}\)
This results in the replacement of \(\ce{H^+}\), which exhibit high conductivity, with \(\ce{NH4^+}\), which have lower conductivity. Solution conductivity decreases.
The conductivity at the equivalence point is the lowest as only \(\ce{Cl^-}\) and \(\ce{NH4^+}\) ions are present which both have low relative conductivities.
Beyond the equivalence point, the excess ammonia added reacts partially with water to form \(\ce{NH4^+}\) and \(\ce{OH^-}\) ions according to the equation:
\(\ce{NH3(aq) + H2O(l) \rightleftharpoons NH4^+(aq) + OH^-(aq)}\)
Although these ions are more conductive than the reactant molecules, the ionisation of ammonia is limited due to the presence of \(\ce{NH4^+}\) ions already in the solution (as per Le Châtelier’s Principle).
Consequently, the conductivity increases only slightly after the equivalence point.

CHEMISTRY, M6 2024 HSC 17 MC

20 mL of a 0.1 mol L\(^{-1}\) solution of an acid is titrated against a 0.1 mol L\(^{-1}\) solution of sodium hydroxide. A graph of pH against the volume of sodium hydroxide for this experiment is shown.

Which of the following acids was used in the titration?

\begin{align*}
\begin{array}{l}
\ & \\
\textbf{A.}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|c|c|c|}
\hline
\quad \textit{Acid}\quad & \quad pK_{a1}\quad & \quad pK_{a2}\quad \\
\hline
1& 4.76 & – \\
\hline
2 & \text{Strong} & – \\
\hline
3 & 1.91 & 6.30 \\
\hline
4 & 4.11 & 9.61 \\
\hline
\end{array}
\end{align*}

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\(C\)

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There are two equivalence points on the graph, so the acid is diprotic and will have two \(pK_a\) values.
The reaction at the first equivalence point is:
\(\ce{H2A -> HA^- + H^+}\)
The reaction at the second equivalence point is:
\(\ce{HA^- -> A^{2-} + H^+}\)
The \(pK_a\) values for each equivalence point will be equal to the pH values at the half-equivalence points.
The first half-equivalence point occurs at 10 mL, where the corresponding pH is 1.91 \(\Rightarrow pK_{a1}\) = 1.91
The second half-equivalence point will occur halfway between the first two equivalence points, at 30 mL
The corresponding pH is 6.30 \(\Rightarrow pK_{a2}\) = 6.30

\(\Rightarrow C\)

♦ Mean mark 41%.
COMMENT: Student’s did well understanding the relationship between the \(pK_a\) values and half-equivalence points.

CHEMISTRY, M6 2023 HSC 9 MC

A titration was performed using two solutions of equal concentration, producing the following titration curve.

Which combination of solutions does the titration curve represent?

Addition of a weak base to a weak acid
Addition of a weak base to a strong acid
Addition of a strong acid to a weak base
Addition of a strong acid to a strong base

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\(C\)

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Weak base acts as a buffer, resisting an immediate decrease in pH upon addition of strong acid

\(\Rightarrow C\)

CHEMISTRY, M6 EQ-Bank 25

The graph shows changes in pH for the titrations of equal volumes of solutions of two monoprotic acids, Acid 1 and Acid 2.

Explain the differences between Acid 1 and Acid 2 in terms of their relative strengths and concentrations. (3 marks)

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Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.
Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.
Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more \(\ce{KOH}\) to neutralise it.

Show Worked Solution

Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.
Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.
Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more \(\ce{KOH}\) to neutralise it.

CHEMISTRY, M6 2019 HSC 24

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.

Explain the shape of the titration curve. (3 marks)

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The equivalence point was reached when a volume of 17.15 mL of barium hydroxide was added.
Calculate the concentration of barium hydroxide (in mol L¯¹), and give a relevant chemical equation. (4 marks)

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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. \(\ce{2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2H2O(l)} \)

\(\ce{[Ba(OH)2]} = 7.653 \times 10^{-3}\ \text{mol L}^{-1}\)

Show Worked Solution

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.

♦ Mean mark (a) 46%.

b. \(\ce{2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2H2O(l)} \)

\(\ce{n(HCl) = c \times V = 1.050 \times 10^{-3} \times 250 = 2.625 \times 10^{-4} mol }\)

\(\ce{n(Ba(OH)2) = 0.5 \times n(HCl) = 1.3125 \times 10^{-4} mol}\)

\(\ce{[Ba(OH)2] = \dfrac{\text{n}}{\text{V}}}=\dfrac{1.3125 \times 10^{-4}}{0.01715}=7.653 \times 10^{-3}\ \text{mol L}^{-1}\)

CHEMISTRY, M6 2019 HSC 6 MC

The diagram represents the titration curve for a reaction between a particular acid a particular base.

Which of the following equations best represents the reaction described by the titration curve?

\(\ce{NH3 (aq) + HCl(aq) to NH4 Cl (aq)}\)
\(\ce{NaOH(aq) + HCl(aq) to NaCl(aq) + H2O(l)}\)
\(\ce{NH3 (aq) + CH3 COOH(aq) to CH3COONH4(aq)}\)
\(\ce{NaOH(aq) + CH3 COOH(aq) to CH3COONa(aq)+H2O (l)}\)

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\(A\)

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The titration curve shows the progress of a titration in which a weak base \(\ce{NH3}\) is reacting with a strong acid \(\ce{CHl}\) .
As the acid is added to the base, the pH of the solution moves lower.

\(\Rightarrow A\)

CHEMISTRY, M6 2019 HSC 5 MC

The diagram represents the titration curve for a reaction between a particular acid and a particular base.

Which indicator would be best for this titration?

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`C`

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The pH range at which isopicramic acid exhibits a colour change includes the point at which the acid and base react in equal amounts (equivalence point), which is at approximately pH 5.
The colour change can be used to identify when the equivalence point has been reached in a titration.

`=>C`

CHEMISTRY, M6 2020 HSC 8 MC

A weak base is titrated with 1.0 mol L\(^{-1}\) aqueous \(\ce{HCI}\). The \(\ce{pH}\) curve is shown.

At which pH value would the solution be most effective as a buffer?

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\(D\)

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The solution would be most effective as a buffer at \(\ce{pH}\) 9 because this corresponds to the half-equivalent point.

\(\Rightarrow D\)

♦♦ Mean mark 37%.

CHEMISTRY, M6 2022 HSC 15 MC

A 25.00 mL sample of 0.1131 `\text{mol}\ text{L}^(-1)` `\text{HCl}(aq)` was titrated with an aqueous ammonia solution. The conductivity of the solution was measured throughout the titration and the results graphed.

What was the concentration of the ammonia solution?

0.0452 `\text{mol}\ text{L}^(-1)`
0.189 `\text{mol}\ text{L}^(-1)`
0.283 `\text{mol}\ text{L}^(-1)`
0.690 `\text{mol}\ text{L}^(-1)`

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`C`

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The equivalence point occurs at 10.0 mL of ammonia.

`text{HCl} (aq) + text{NH}_3 (aq) → text{NH}_4 text{Cl} (aq)`

`text{n(HCl)}`	`= text{c} xx text{V}`
	`= 0.1131 xx 0.02500`
	`= 2.828 xx text{10}^(−3) text{mol}`

`text{n(NH}_3 )= text{n(HCl)}= 2.828 xx text{10}^(−3) text{mol}`

`:.\ text{[NH}_3]= text{n} / text{V}= [2.828 xx text{10}^(−3)] / 0.0100= 0.283 text{mol L}^(–1)`

`=> C`

CHEMISTRY, M6 2021 HSC 16 MC

This titration curve is produced when an acid is titrated with a sodium hydroxide solution of the same concentration.

How many acidic protons does this acid possess?

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`B`

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From the graph there are 2 equivalence points (at ~ 10 mL and 20 mL), indicating that the acid is diprotic.
i.e. it contains 2 acidic protons.

`=> B`

♦ Mean mark 39%.