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Chapter 14: Problem 112

Hydrochloric acid is added to an aqueous solution of silver nitrate. (a) What precipitate forms? (b) The equilibrium concentration of \(\mathrm{Cl}^{-}(a q)\) is \(2.0 \times 10^{-3} \mathrm{M}\). What is the equilibrium concentration of \(\mathrm{Ag}^{+}(a q) ?\)

Short Answer

Expert verified
(a) The precipitate formed is AgCl (silver chloride). (b) The equilibrium concentration of \(\mathrm{Ag}^{+}(a q)\) is \(9.0 \times 10^{-8} \mathrm{M}\).

Step by step solution

01

(a) Balanced chemical equation

Write the balanced chemical equation for the reaction between hydrochloric acid and silver nitrate. \[ \mathrm{HCl(aq)}+\mathrm{AgNO_{3}(aq)} \rightarrow \mathrm{AgCl(s)}+\mathrm{HNO_{3}(aq)} \] Here, AgCl is the precipitate that forms.
02

(b) Solubility product constant

We will now consider the solubility product constant of AgCl, denoted by \(K_{sp}\), which represents the equilibrium between the dissolved ions and the solid: \[ \mathrm{AgCl(s)} \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \] The solubility product constant expression can be written as: \[ K_{sp} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]
03

(b) Given concentration of Cl^- ions

The problem provides the equilibrium concentration of \(\mathrm{Cl}^{-}(a q)\) as \(2.0 \times 10^{-3} \mathrm{M}\).
04

(b) Finding Ksp value for AgCl

Find the \(K_{sp}\) value for AgCl from a reference table or other source. For example, the \(K_{sp}\) value for AgCl is \(1.8 \times 10^{-10}\).
05

(b) Calculating equilibrium concentration of Ag+ ions

Using the \(K_{sp}\) expression and the given concentration of \(\mathrm{Cl}^{-}(a q)\), we can find the equilibrium concentration of \(\mathrm{Ag}^{+}(a q)\): \[ \begin{aligned} K_{sp}&=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \\ 1.8 \times 10^{-10}&=[\mathrm{Ag}^{+}](2.0 \times 10^{-3} \mathrm{M}) \\ \end{aligned} \] Now, calculate the equilibrium concentration of \(\mathrm{Ag}^{+}(a q)\): \[ \begin{aligned} [\mathrm{Ag}^{+}] &= \frac{1.8 \times 10^{-10}}{2.0 \times 10^{-3} \mathrm{M}} \\ [\mathrm{Ag}^{+}] &= 9.0 \times 10^{-8} \mathrm{M} \\ \end{aligned} \] Therefore, the equilibrium concentration of \(\mathrm{Ag}^{+}(a q)\) is \(9.0 \times 10^{-8} \mathrm{M}\).

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Most popular questions from this chapter

Write the equilibrium constant expression for (a) \(2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) \(+6 \mathrm{HCl}(g)\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftarrows 2 \mathrm{Fe}(l)+3 \mathrm{CO}_{2}(g)\) (c) \(\mathrm{PbSO}_{4}(s) \rightleftarrows \mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) (d) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{CO}_{3}(a q)\)

The solubility of iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}\), is \(1.43 \times 10^{-3} \mathrm{~g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) (a) Write a balanced equation for the solubility equilibrium. (b) Write the expression for \(K_{\mathrm{sp}}\) and calculate its value. (c) \(\mathrm{A} 60.0 \mathrm{~mL}\) sample of \(3.00 \times 10^{-3} \mathrm{M} \mathrm{FeSO}_{4}\) solution is added to \(60.0 \mathrm{~mL}\) of \(4.00 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) solution. Does a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{2}\) form? (Hint: Use the concentration values for \(\mathrm{Fe}^{2+}\) and \(\mathrm{OH}^{-}\) and plug them into the \(K_{\mathrm{sp}}\) expression. If the value you get is larger than \(K_{\mathrm{sp}}\) ' precipitation will occur.)

What do we call a reaction when we say that it can proceed in both the forward and reverse directions? In principle, which chemical reactions can proceed in both directions?

Cooling an exothermic reaction for which \(K_{\text {eq }}\) is very low shifts the reaction to the right, so that more product is formed, but there is a trade-off. What is the downside of cooling such a reaction, as far as forming product is concerned?

What does the "equi" portion of the word equilibrium refer to?
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