Calculate the pH of each of the following solutions: (a) A 0.025 M HNO3 solution (b) A 0.10 M solution of NaOH (c) A solution prepared by dissolving 0.185 g of slaked lime [Ca(OH)2] in enough water to produce 0.50 L.

Question

Calculate the pH of each of the following solutions:

(a) A 0.025 M [latex]HNO_{3}[/latex] solution

(b) A 0.10 M solution of NaOH

(c) A solution prepared by dissolving 0.185 g of slaked lime [latex][Ca(OH)_{2}][/latex] in enough water to produce 0.50 L.

STRATEGY

Write the dissociation reaction for each acid or base and use the stoichiometry of the balanced equation to find the [latex][H_{3}O^{+}][/latex] or [latex][OH^{-}][/latex] in solution. Strong acids and strong bases are essentially 100% dissociated in aqueous solution. The pH equals the negative log of the [latex][H_{3}O^{+}][/latex]. For solutions of strong bases, [latex][H_{3}O^{+}][/latex] can be found from [latex][OH^{-}][/latex] using the expression for [latex]K_{w}, [H_{3}O^{+}] = K_{w}/[OH^{-}][/latex].

Accepted Answer

(a) Because strong acids are completely dissociated, [latex][H_{3}O^{+}][/latex] equals the initial concentration of [latex]HNO_{3}[/latex].

[latex]HNO_{3}(aq)+H_{2}O(l)\overset{100\%}{\longrightarrow}H_{3}O^{+}(aq)+NO_{3} ^{-}(aq)[/latex]

[latex]pH=-log [H_{3}O^{+}]=-log(2.5 imes 10^{-2})=1.60[/latex]

(b) Because strong bases are completely dissociated, [latex][OH^{-}][/latex] equals the initial concentration of NaOH (0.10 M).

[latex]NaOH(s)\overset{H_{2}O}{\longrightarrow}NaOH(aq)\overset{100\%}{\longrightarrow}Na^{+}(aq)+OH^{-}(aq)[/latex]

[latex][H_{3}O^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{1.0 imes 10^{-14}}{0.10}=1.0 imes 10^{-13} M[/latex]

[latex]pH=-log (1.0 imes 10^{-13})=13.00[/latex]

(c) First calculate the number of moles of [latex]Ca(OH)_{2}[/latex] dissolved from the given mass of [latex]Ca(OH)_{2}[/latex] and its molar mass (74.1 g/mol).

[latex]0.185 g Ca(OH)_{2} imes\frac{1 mol Ca(OH)_{2}}{74.1 g Ca(OH)_{2}}=2.50 imes 10^{-3} mol[/latex]

Divide the number of moles by volume to find the molar concentration of [latex]Ca(OH)_{2}[/latex].

[latex][Ca(OH)_{2}]=\frac{2.50 imes 10^{-3} mol Ca(OH)_{2}}{0.50 L}=5.0 imes 10^{-3} M[/latex]

Because slaked lime is a strong base and completely dissociated, it provides [latex]2 OH^{-}[/latex] ions per [latex]Ca(OH)_{2}[/latex] formula unit. Therefore,[latex][OH^{-}][/latex]=2(0.0050 M) = 0.010 M.

[latex]Ca(OH)_{2}(s)\overset{H_{2}O}{\longrightarrow}Ca(OH)_{2}(aq)\overset{100\%}{\longrightarrow}Ca^{2+}(aq)+2 OH^{-}(aq) [/latex]

[latex][H_{3}O^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{1.0 imes 10^{-14}}{0.010}=1.0 imes 10^{-12} M[/latex]

[latex]pH=-log (1.0 imes 10^{-12})=12.00[/latex]

IDENTIFY
Known	Unknown
Concentration of strong acid or base	pH

Chapter 15

Q. 15.8

Q. 15.8

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