Question 15.8: Calculate the pH of each of the following solutions: (a) A 0......
Calculate the pH of each of the following solutions:
(a) A 0.025 M HNO_{3} 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.
STRATEGY
Write the dissociation reaction for each acid or base and use the stoichiometry of the balanced equation to find the [H_{3}O^{+}] or [OH^{-}] in solution. Strong acids and strong bases are essentially 100% dissociated in aqueous solution. The pH equals the negative log of the [H_{3}O^{+}]. For solutions of strong bases, [H_{3}O^{+}] can be found from [OH^{-}] using the expression for K_{w}, [H_{3}O^{+}] = K_{w}/[OH^{-}].
| IDENTIFY | |
| Known | Unknown |
| Concentration of strong acid or base | pH |
(a) Because strong acids are completely dissociated, [H_{3}O^{+}] equals the initial concentration of HNO_{3}.
HNO_{3}(aq)+H_{2}O(l)\overset{100\%}{\longrightarrow}H_{3}O^{+}(aq)+NO_{3} ^{-}(aq)pH=-log [H_{3}O^{+}]=-log(2.5\times 10^{-2})=1.60
(b) Because strong bases are completely dissociated, [OH^{-}] equals the initial concentration of NaOH (0.10 M).
NaOH(s)\overset{H_{2}O}{\longrightarrow}NaOH(aq)\overset{100\%}{\longrightarrow}Na^{+}(aq)+OH^{-}(aq)[H_{3}O^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{0.10}=1.0\times 10^{-13} M
pH=-log (1.0\times 10^{-13})=13.00
(c) First calculate the number of moles of Ca(OH)_{2} dissolved from the given mass of Ca(OH)_{2} and its molar mass (74.1 g/mol).
0.185 g Ca(OH)_{2}\times\frac{1 mol Ca(OH)_{2}}{74.1 g Ca(OH)_{2}}=2.50\times 10^{-3} molDivide the number of moles by volume to find the molar concentration of Ca(OH)_{2}.
[Ca(OH)_{2}]=\frac{2.50\times 10^{-3} mol Ca(OH)_{2}}{0.50 L}=5.0\times 10^{-3} MBecause slaked lime is a strong base and completely dissociated, it provides 2 OH^{-} ions per Ca(OH)_{2} formula unit. Therefore,[OH^{-}]=2(0.0050 M) = 0.010 M.
Ca(OH)_{2}(s)\overset{H_{2}O}{\longrightarrow}Ca(OH)_{2}(aq)\overset{100\%}{\longrightarrow}Ca^{2+}(aq)+2 OH^{-}(aq)[H_{3}O^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{0.010}=1.0\times 10^{-12} M
pH=-log (1.0\times 10^{-12})=12.00