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Chapter 17: Problem 54

(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate $K_{s p} .(\mathbf{b})\( If \)0.0490 \mathrm{~g}\( of \)\mathrm{AgIO}_{3}$ dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Short Answer

Expert verified
The solubility-product constant (\(K_{sp}\)) for \(\mathrm{PbBr}_{2}\) is \(4.0 \times 10^{-6}\). The \(K_{sp}\) for \(\mathrm{AgIO}_{3}\) is \(4.44 \times 10^{-8}\). The pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is 12.51.

Step by step solution

01

Write the balanced chemical equation

The dissolution of \(\mathrm{PbBr}_{2}\) in water can be represented by the following balanced equation: \[\mathrm{PbBr}_{2}(s)\rightleftharpoons \mathrm{Pb^{2+}}(aq)+2 \mathrm{Br^-}(aq)\]
02

Write the expression for solubility-product constant

The solubility-product constant expression is: \[K_{sp}=[\mathrm{Pb}^{2+}][\mathrm{Br}^{-}]^{2}\]
03

Calculate the concentrations of the ions

Molar solubility of \(\mathrm{PbBr}_{2}\) is given as \(1.0 \times 10^{-2} \mathrm{M}\). Since the stoichiometry is 1:2 in the balanced equation, this means that the concentration of \(\mathrm{Br}^{-}\) is twice the concentration of \(\mathrm{Pb}^{2+}\). So, the concentrations are: [\(\mathrm{Pb^{2+}}\)] \(= 1.0 \times 10^{-2} \mathrm{M}\) [\(\mathrm{Br^-}\)] \(= 2(1.0 \times 10^{-2}) \mathrm{M} = 2.0 \times 10^{-2} \mathrm{M}\)
04

Calculate the solubility-product constant

Now, we can plug these values into the \(K_{sp}\) expression: \[K_{sp}=(1.0 \times 10^{-2})(2.0 \times 10^{-2})^{2} = 4.0 \times 10^{-6}\] Thus, the \(K_{sp}\) for \(\mathrm{PbBr}_{2}\) is \(4.0 \times 10^{-6}\). (b) Calculate the solubility-product constant for \(\mathrm{AgIO}_{3}\).
05

Write the balanced chemical equation

The dissolution of \(\mathrm{AgIO}_{3}\) in water can be represented by the following balanced equation: \[\mathrm{AgIO}_{3}(s)\rightleftharpoons \mathrm{Ag^{+}}(aq)+\mathrm{IO}_{3}^{-}(aq)\]
06

Write the expression for solubility-product constant

The solubility-product constant expression is: \[K_{sp}=[\mathrm{Ag}^{+}][\mathrm{IO}_{3}^{-}]\]
07

Calculate the concentrations of the ions

Given that \(0.049 \mathrm{g}\) of \(\mathrm{AgIO}_{3}\) dissolve per liter of solution, the molar solubility of \(\mathrm{AgIO}_{3}\) can be calculated using the molar mass of \(\mathrm{AgIO}_{3}\), which is approximately \(232 \mathrm{g/mol}\). Molar solubility of \(\mathrm{AgIO}_{3}\): \(\frac{0.049\;\mathrm{g}}{232\;\mathrm{g/mol}} = 2.11 \times 10^{-4} \;\mathrm{M}\) Since the stoichiometry is 1:1, we have: [\(\mathrm{Ag^{+}}\)] \(= 2.11 \times 10^{-4} \;\mathrm{M}\) [\(\mathrm{IO}_{3}^{-}\)] \(= 2.11 \times 10^{-4} \;\mathrm{M}\)
08

Calculate the solubility-product constant

Now, we can plug these values into the \(K_{sp}\) expression: \[K_{sp}=(2.11 \times 10^{-4})(2.11 \times 10^{-4}) = 4.44 \times 10^{-8}\] Thus, the \(K_{sp}\) for \(\mathrm{AgIO}_{3}\) is \(4.44 \times 10^{-8}\). (c) Calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).
09

Write the balanced chemical equation

The dissolution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water can be represented by the following balanced equation: \[\mathrm{Ca}(\mathrm{OH})_{2}(s) \leftrightarrows \mathrm{Ca}^{2+}(aq) + 2 \mathrm{OH}^{-}(aq)\]
10

Write the expression for solubility-product constant

The solubility-product constant expression is: \[K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{OH}^{-}]^{2}\]
11

Use the appendix to find the \(K_{sp}\) value for \(\mathrm{Ca}(\mathrm{OH})_{2}\)

According to Appendix D, the \(K_{sp}\) value for \(\mathrm{Ca}(\mathrm{OH})_{2}\) at \(25^{\circ} \mathrm{C}\) is \(5.02 \times 10^{-6}\).
12

Calculate the concentration of \(\mathrm{OH}^-\)

Let the molar solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) be \(s\). Then we have: [\(\mathrm{Ca^{2+}}\)] \(=s\) [\(\mathrm{OH}^{-}\)] \(=2s\) Plugging these into the \(K_{sp}\) expression, we get: \(5.02 \times 10^{-6}=s(2s)^{2}\) Solving for \(s\), we get \(s=1.61 \times 10^{-2} \;\mathrm{M}\). Since \(\mathrm{OH}^{-}\) ion concentration is twice of \(s\), we have: [\(\mathrm{OH}^{-}\)] \(=2(1.61 \times 10^{-2}) = 3.22 \times 10^{-2} \;\mathrm{M}\)
13

Calculate the \(pOH\) and then find the pH

Now, we can calculate the \(pOH\): \(pOH=-\log([\mathrm{OH}^{-}])=-\log(3.22 \times 10^{-2})=1.49\) Using the relationship between \(pOH\) and \(pH\) (\(pH + pOH = 14\)), we can calculate the pH: \(pH=14-pOH=14-1.49=12.51\) Thus, the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is 12.51.

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Most popular questions from this chapter

(a) Calculate the pH of a buffer that is \(0.150 \mathrm{M}\) in lactic acid and \(0.120 M\) in sodium lactate. (b) Calculate the pH of a buffer formed by mixing \(75 \mathrm{~mL}\) of \(0.150 \mathrm{M}\) lactic acid with \(25 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) sodium lactate.

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the Chemistry and Life box on page 790 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), whose $K_{s p}=1.0 \times 10^{-60}$ (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Consider the titration of \(30.0 \mathrm{~mL}\) of $0.050 \mathrm{M} \mathrm{NH}_{3}\( with \)0.025 M$ HCl. Calculate the pH after the following volumes of titrant have been added: $(\mathbf{a}) 0 \mathrm{~mL},(\mathbf{b}) 20.0 \mathrm{~mL},\( (c) 59.0 \)\mathrm{mL},(\mathbf{d}) 60.0 \mathrm{~mL},(\mathbf{e}) 61.0 \mathrm{~mL},(\mathbf{f}) 65.0 \mathrm{~mL} .$

A buffer contains 0.30 mol of propionic acid $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\( and \)0.25 \mathrm{~mol}$ of potassium propionate \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOK}\right)\) in 1.80 L. (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of \(0.10 \mathrm{~mol}\) of \(\mathrm{NaOH}\) ? \((\mathbf{c})\) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

A sample of \(500 \mathrm{mg}\) of an unknown monoprotic acid was dissolved in \(50.0 \mathrm{~mL}\) of water and titrated with \(0.200 \mathrm{M}\) KOH. The acid required \(20.60 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(10.30 \mathrm{~mL}\) of base had been added in the titration, the pH was found to be 4.20 . What is the \(\mathrm{p} K_{a}\) for the unknown acid?
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