# 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
Step-by-Step
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(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).

$0.185 g Ca(OH)_{2}\times\frac{1 mol Ca(OH)_{2}}{74.1 g Ca(OH)_{2}}=2.50\times 10^{-3} mol$

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

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

Because 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.

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

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

$pH=-log (1.0\times 10^{-12})=12.00$

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