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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 [Ca虏鈦篯[SO鈧劼测伝] and [Ag鈦篯[SO鈧劼测伝] 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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant (Ksp)
The solubility product constant, or \(K_{sp}\), is a crucial concept. It measures the solubility of a compound in water. This constant is specific to each compound at a given temperature.When representing \(K_{sp}\), we look at the concentrations of the ions in a saturated solution. For example, in the case of \(CaSO_{4}\), \(K_{sp}\) would be expressed as:\[K_{sp} = [Ca^{2+}][SO_{4}^{2-}]\]The idea behind \(K_{sp}\) is simple:
  • If the ion product of the solution is less than \(K_{sp}\), the solution is unsaturated, and no precipitate forms.
  • If the ion product equals \(K_{sp}\), the solution is saturated and on the verge of forming a precipitate.
  • If the ion product exceeds \(K_{sp}\), the solution is supersaturated, and a precipitate will form.
Understanding \(K_{sp}\) helps predict if a reaction will form a precipitate, which is a solid that separates from the liquid solution. By knowing the \(K_{sp}\), chemists can anticipate the behavior of different compounds when mixed.
Ion Product
The ion product, denoted as \(Q\), represents the product of the concentrations of the ions present in a solution at any given time. It's similar to the solubility product constant \(K_{sp}\), but unlike \(K_{sp}\), it's calculated under non-equilibrium conditions.For instance, in determining whether \(CaSO_{4}\) or \(Ag_2SO_{4}\) will precipitate first, we need to calculate the ion product for each.
This is done as the sulfate ions \(([SO_4^{2-}])\)from \(Na_2SO_4\) are added to the solution:
  • The ion product for \(CaSO_4\) is \(Q = [Ca^{2+}][SO_{4}^{2-}]\).
  • The ion product for \(Ag_2SO_4\) is \(Q = [Ag^{+}]^2[SO_{4}^{2-}]\).
When the ion product \(Q\) of a compound in a solution exceeds its \(K_{sp}\), a precipitate begins to form. Comparing \(Q\) to \(K_{sp}\) allows chemists to tell whether conditions are just right for a solid to formor whether more ions must dissolve before precipitation occurs.
Precipitation Reaction
When a precipitation reaction occurs, a solid is formed from two substances in a solution. This process depends on the interaction between ions and often involves a shift from a dissolved to a solid state.In our example, as \(Na_2SO_4\) is added to a solution containing \(Ca^{2+}\) and \(Ag^{+}\), we are interested in which will form a precipitate first.Precipitation occurs when:
  • The concentration of ions surpasses the \(K_{sp}\), resulting in a shift from dissolved ions back to a solid.
  • Conditions favor the formation of a stable solid structure.
In chemical terms,\(CaSO_4\) forms a precipitate first when enough \(SO_4^{2-}\) ions have been added to satisfy its ion product reaching > \(K_{sp}\). Precipitates generally form because the solid state is more stable than the high concentration of dissolved ions. This stability drives the reaction, resulting in a solid formation.

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Most popular questions from this chapter

A sample of \(7.5 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) gas at \(22^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50-L solution of \(0.40 \mathrm{M}\) HCl. Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{~L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}(a q)\) and \(0.010 \mathrm{M}\) in \(\mathrm{Sr}^{2+}(a q) .(\mathbf{a}) \mathrm{What}\) concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4}: K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}:\) \(K_{s p}=3.2 \times 10^{-7} .\) ) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) when the second cation begins to precipitate?

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?(\mathbf{b})\) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

Assume that \(30.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of a weak base \(\mathrm{B}\) that accepts one proton is titrated with a \(0.10 \mathrm{M}\) solution of the monoprotic strong acid HA. (a) How many moles of HA have been added at the equivalence point? (b) What is the predominant form of B at the equivalence point? (a) Is the \(\mathrm{pH} 7\), less than 7 , or more than 7 at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either \(\mathrm{M}^{+}\) or \(\mathrm{A}^{-} .\) (b) Common ions alter the equilibrium constant for the reaction of an ionic solid with water. \((\mathbf{c})\) The common-ion effect does not apply to unusual ions like \(\mathrm{SO}_{3}^{2-} .(\mathbf{d})\) The solubility of a salt MA is affected equally by the addition of either \(\mathrm{A}^{-}\) or a noncommon ion.
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