Which of the following metal carbonates has the highest molar solubility?
- Calcium carbonate
- Copper`text{(II)}` carbonate
- Iron`text{(II)}` carbonate
- Lead`text{(II)}` carbonate
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Which of the following metal carbonates has the highest molar solubility?
`A`
→ Calcium carbonate has the greatest solubility product among the listed metal carbonates.
→ Since all of these metal carbonates release the same amount of ions in solutions, their solubility products can be directly compared to determine which substance is most soluble.
`=>A`
Equal volumes of two 0.04 mol L ¯1 solutions were mixed together.
Which pair of solutions would give the greatest mass of precipitate?
`B`
\(\ce{Ba(OH)2(aq) + MgSO4(aq) → BaSO4(s) + Mg(OH)2(s)}\)
→ Reaction B produces 2 molecules of precipitate
→ Reactions A and D produce 1 molecule of precipitate each
→ Reaction C does not produce a precipitate.
`=> B`
Sodium hypochlorite `\text{NaOCl}` is the active ingredient in pool chlorine. It completely dissolves in water to produce the hypochlorite ion `(\text{OCl}^(-))`, which undergoes hydrolysis according to the following equilibrium.
\( \ce{OCl-(aq) + H2O(l) \rightleftharpoons HOCl(aq) + OH-(aq)} \)
The equilibrium constant for this reaction at 25°C is `3.33 xx 10^(-7)`.
For pool chlorine to be effective the pH is maintained by a different buffer at 7.5 and the hypochlorous acid `(\text{HOCl})` concentration should be `1.3 xx 10^(-4)` mol L ¯1.
Calculate the volume of 2.0 mol L ¯1 sodium hypochlorite solution that needs to be added to a 1.00 × 104 L pool to meet the required conditions. (4 marks)
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`1.3\ text{L}`
`text{pOH}_(eq) = 14.00 − 7.5 = 6.5`
`[text{OH}^–]_(eq) = 10^-text{pOH} = 10^(−6.5)\ text{mol L}^(–1)`
`text{K}_(eq) ` | `=[[text{HOCl}]_(eq) [text{OH}^–]_(eq)] / [text{OCl}^–]_(eq)` | |
`3.33 xx 10^(−7)` | `= [(1.3 × 10^(−4)) xx (10^(−6.5))] / [[text(OCl)^–]_(eq)]` | |
`[text(OCl)^–]_(eq)` | `=[(1.3 × 10^(−4)) xx (10^(−6.5))]/(3.33 xx 10^(−7))` | |
`= 1.246 xx 10^(−4)\ text{mol L}^(–1)` |
\begin{array} {|l|c|c|c|}
\hline & \text{OCl}^– & \text{HOCl} & \text{OH}^– \\
\hline \text{Initial} & x & 0 & – \\
\hline \text{Change} & – 1.3 \times 10^{–4} & +1.3 \times 10^{−4} & – \\
\hline \text{Equilibrium} & x − 1.3 \times 10^{−4} & 1.3 \times 10^{−4} & 10^{−6.5} \\
\hline \end{array}
`x−1.3 xx 10^(−4)` | `= 1.246 xx 10^(−4)` | |
`x` | `= 2.546 xx 10^(−4)` |
`[text{OCl}^–]_i = 2.55 xx 10^(−4)\ text{mol L}^(–1)\ \ text{(3 s.f.)}`
\( \ce{[NaOCl] \times V(NaOC)_{req}}\) | \( \ce{= [OCl-](pool) \times V(pool)} \) | |
\( \ce{V(NaCl)_{req}} \) | `=(2.55 xx 10^(−4) xx 10^4)/2` | |
`=1.3\ text{L (2 s.f.)}` |
Silver ions form the following complex with ammonia solution.
\( \ce{Ag+(aq) + 2NH3(aq) \rightleftharpoons [Ag(NH3)2]+(aq)}\)
The equilibrium constant is `1.6 × 10^(7)` at 25°C.
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a. The method is not suitable.
→ Adding \( \ce{NaI}\) would cause the \( \ce{I-}\) ions to precipitate with the \( \ce{Ag+}\) ions to form \( \ce{AgI}\)
\( \ce{AgI(s) \rightleftharpoons Ag+(aq) + I- (aq)}\)
→ As a result, this would decrease \( \ce{[Ag+]}\), and disturb the equilibrium.
→ According to Le Chatelier’s Principle, the equilibrium will shift to the right in an attempt to counteract the change and increase \( \ce{[Ag+]}\).
\( \ce{Ag+(aq) + 2NH3(aq) \rightleftharpoons [Ag(NH3)2]+(aq)}\)
→ As a result, \( \ce{[Ag(NH3)2]+}\) would shift to the left and increase \( \ce{[Ag+]} \).
b.
\[\ce {K_{eq}=\frac{[[Ag(NH3)2]+]}{[Ag+][NH3]^2}} \]
\[\ce {[[Ag(NH3)2]+] = \frac{99.99%}{0.010%} \times [Ag+] \ \ \ …\ (1)}\]
Substitute (1) into \(\ce {K_{eq}:}\)
`1.6 xx 10^7` | `= [(99.99%) xx [text{Ag}^+ ]] / [(0.010%) xx [text{Ag}^+ ][text{NH}_3 ]^2]` |
`= [99.99%] / [0.010% [text{NH}_3 ]^2]` | |
`[text{NH}_3 ]^2` | `=(99.99%)/(1.6 xx 10^7 xx 0.010%)` |
`[text{NH}_3 ]` | `=sqrt((99.99%)/(1.6 xx 10^7 xx 0.010%))` |
`= 0.025 text{mol L}^-1` |
Therefore, the concentration of \(\ce {NH3}\) at equilibrium is 0.025 mol L¯1.
a. The method is not suitable.
→ Adding \( \ce{NaI}\) would cause the \( \ce{I-}\) ions to precipitate with the \( \ce{Ag+}\) ions to form \( \ce{AgI}\)
\( \ce{AgI(s) \rightleftharpoons Ag+(aq) + I- (aq)}\)
→ As a result, this would decrease \( \ce{[Ag+]}\), and disturb the equilibrium.
→ According to Le Chatelier’s Principle, the equilibrium will shift to the right in an attempt to counteract the change and increase \( \ce{[Ag+]}\).
\( \ce{Ag+(aq) + 2NH3(aq) \rightleftharpoons [Ag(NH3)2]+(aq)}\)
→ As a result, \( \ce{[Ag(NH3)2]+}\) would shift to the left and increase \( \ce{[Ag+]} \).
b.
\[\ce {K_{eq}=\frac{[[Ag(NH3)2]+]}{[Ag+][NH3]^2}} \]
\[\ce {[[Ag(NH3)2]+] = \frac{99.99%}{0.010%} \times [Ag+] \ \ \ …\ (1)}\]
Substitute (1) into \(\ce {K_{eq}:}\)
`1.6 xx 10^7` | `= [(99.99%) xx [text{Ag}^+ ]] / [(0.010%) xx [text{Ag}^+ ][text{NH}_3 ]^2]` |
`= [99.99%] / [0.010% [text{NH}_3 ]^2]` | |
`[text{NH}_3 ]^2` | `=(99.99%)/(1.6 xx 10^7 xx 0.010%)` |
`[text{NH}_3 ]` | `=sqrt((99.99%)/(1.6 xx 10^7 xx 0.010%))` |
`= 0.025 text{mol L}^-1` |
Therefore, the concentration of \(\ce {NH3}\) at equilibrium is 0.025 mol L¯1.
What is the molar solubility of iron(`text{II}`) hydroxide?
`A`
`text{Fe(OH)}_2 (s) ⇌ text{Fe}^(2+) (aq) + 2 text{OH}^– (aq)`
Solids are not included in the `K_(sp)` expression
\begin{array} {|l|c|c|}
\hline & \text{Fe}^{2+} & \ \ \text{OH}^– \ \ \\
\hline \text{Initial} & 0 & 0 \\
\hline \text{Change} & + x & + 2x \\
\hline \text{Equilibrium} & x & 2x \\
\hline \end{array}
`text{K}_(sp)` | `= [text{Fe}^(2+)][text{OH}^–]^2` |
`4.87 xx 10^(−17) ` | `= x xx (2x)^2` |
`4.87 xx 10^(−17)` | `= 4x^3` |
`:.x` | `= root3((4.87 xx 10^(−17))/4)` |
`=2.30 xx 10^(−6) text{mol L}^(–1)` |
`=> A`