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Chapter 17: Problem 76

A \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution is slowly added to \(10.0 \mathrm{~mL}\) of a solution that is \(0.20 M\) in \(\mathrm{Ca}^{2+}\) and \(0.30 \mathrm{M}\) in \(\mathrm{Ag}^{+} .\) (a) Which compound will precipitate first: \(\operatorname{CaSO}_{4}\left(K_{s p}=2.4 \times 10^{-5}\right)\) or $\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{s p}=1.5 \times 10^{-5}\right) ?(\mathbf{b})\( How much \)\mathrm{Na}_{2} \mathrm{SO}_{4}$ solution must be added to initiate the precipitation?

Short Answer

Expert verified
(a) \(\mathrm{CaSO}_{4}\) will precipitate first. (b) Approximately \(83.3 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added to initiate the precipitation of \(\mathrm{CaSO}_{4}\).

Step by step solution

01

Calculate the initial ion product for both compounds

First, we need to calculate the initial ion product [Ca²⁺][SO₄²⁻] and [Ag⁺][SO₄²⁻] without considering the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution. For \(\operatorname{CaSO}_{4}\): \[[Ca^{2+}] = 0.20 M\] \[[SO_{4}^{2-}] = 0\] So the initial ion product for \(\operatorname{CaSO}_{4}\) is: \[((0.20)(0))=0\] For \(\mathrm{Ag_{2}SO}_{4}\): \[[Ag^{+}] = 0.30 M\] \[[SO_{4}^{2-}] = 0\] So the initial ion product for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) is: \[((0.30)(0))=0\]
02

Calculate the Q value as the solution is added

When the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution is added, the concentration of \(SO_{4}^{2-}\) ions increases. Let the volume of added solution be \(V\). For \(\operatorname{CaSO}_{4}\): \[[SO_{4}^{2-}] = (1.0M \cdot V)/(V+10)\] So the ion product (Q) after adding the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution is: \[Q_{CaSO_4} = [Ca^{2+}][SO_{4}^{2-}] = (0.20)((1.0M \cdot V)/(V+10))\] For \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\): \[[SO_{4}^{2-}] = (1.0M \cdot V)/(V+10)\] So the ion product (Q) for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\), after addition of the \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution is: \[Q_{Ag_2SO_4} = [Ag^{+}][SO_{4}^{2-}] = (0.30)((1.0M \cdot V)/(V+10))\]
03

Determine the first precipitate

Whichever compound precipitates first will have its Q value reach the \(K_{sp}\) first. Since both Q expressions have the same \([SO_{4}^{2-}]\) term, we can determine which reaches their \(K_{sp}\) first by comparing the ratio of Q/Ksp for each compound. Solve for "V" to determine which compound has a smaller volume required to start precipitation. For \(\operatorname{CaSO}_{4}\): \[Q_{CaSO_4}/K_{sp}=(0.20)((1.0\cdot V)/(V+10))/(2.4 \times 10^{-5})=1\] For \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\): \[Q_{Ag_2SO_4}/K_{sp}=(0.30)((1.0\cdot V)/(V+10))/(1.5 \times 10^{-5})=1\] Solve for V for both equations.
04

Calculate the Volumes

For \(\operatorname{CaSO}_{4}\): \[V_{CaSO_4} = \frac{0.20 \cdot 10 + 2.4 \times 10^{-5} \cdot V}{2.4 \times 10^{-5}} - 10\] \[V_{CaSO_4} \approx 83.3 mL\] For \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\): \[V_{Ag_2SO_4} = \frac{0.30 \cdot 10 + 1.5 \times 10^{-5} \cdot V}{1.5 \times 10^{-5}} - 10\] \[V_{Ag_2SO_4} \approx 199.3 mL\] Since \(V_{CaSO_4} < V_{Ag_2SO_4}\), \(\mathrm{CaSO}_{4}\) will precipitate first.
05

Answer the question

(a) The compound that precipitates first is \(\mathrm{CaSO}_{4}\). (b) To initiate the precipitation of \(\mathrm{CaSO}_{4}\), approximately \(83.3 mL\) of \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added.

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