Finding the [OH-] and pH of a Strong Base Solution What is the OH- concentration and pH in each solution? (a) 0.225 M KOH (b) 0.0015 M Sr(OH)2

Question

Finding the [OH[latex]^{-}[/latex]] and pH of a Strong Base Solution

What is the OH[latex]^{-}[/latex] concentration and pH in each solution?

(a) 0.225 M KOH (b) 0.0015 M Sr(OH)[latex]_{2}[/latex]

Accepted Answer

(a) Since KOH is a strong base, it completely dissociates into K[latex]^{+}[/latex] and OH[latex]^{-}[/latex] in solution. The concentration of OH[latex]^{-}[/latex] is therefore, the same as the given concentration of KOH.
Use this concentration and K[latex]_{w}[/latex] to find [ [latex]H_{3}O^{+}[/latex] ].
Then substitute [ [latex]H_{3}O^{+}[/latex] ] into the pH expression to find the pH.

KOH(aq) → [latex]K^{+}(aq) + OH^{-}(aq)[/latex]
[OH[latex]^{-}[/latex]] = 0.225 M
[latex][H_{3}O^{+}][OH^{-}] = K_{w} = 1.00 imes 10^{-14}[/latex]
[ [latex]H_{3}O^{+}[/latex] ](0.225) = 1.00 × 10[latex]^{-14}[/latex]
[ [latex]H_{3}O^{+}[/latex] ] = 4.44 × 10[latex]^{-14}[/latex] M
pH = -log[ [latex]H_{3}O^{+}[/latex] ]
= -log(4.44 × 10[latex]^{-14}[/latex])
= 13.353

(b) Since Sr(OH)[latex]_{2}[/latex] is a strong base, it completely dissociates into 1 mol of Sr[latex]^{2+}[/latex] and 2 mol of OH[latex]^{-}[/latex] in solution. The concentration of OH[latex]^{-}[/latex] is therefore, twice the given concentration of Sr(OH)[latex]_{2}[/latex].
Use this concentration and K[latex]_{w}[/latex] to find [ [latex]H_{3}O^{+}[/latex] ].
Substitute [ [latex]H_{3}O^{+}[/latex] ] into the pH expression to find the pH.

[latex]Sr^{2+}(aq) → Sr^{2+}(aq) + 2 OH^{-}(aq)[/latex]
[OH[latex]^{-}[/latex]] = 2(0.0015) M
= 0.0030 M
[latex][H_{3}O^{+}][OH^{-}] = K_{w} = 1.0 imes 10^{-14}[/latex]
[ [latex]H_{3}O^{+}[/latex] ](0.0030) = 1.0 × 10[latex]^{-14}[/latex]
[latex][H_{3}O^{+}] = 3.3 imes 10^{-12}[/latex] M
pH = -log[ [latex]H_{3}O^{+}[/latex] ]
= -log(3.3 × 10[latex]^{-12}[/latex])
= 11.48

(a) Since KOH is a strong base, it completely dissociates into K $^{+}$ and OH $^{-}$ in solution. The concentration of OH $^{-}$ is therefore, the same as the given concentration of KOH. Use this concentration and K $_{w}$ to find [ $H_{3}O^{+}$ ]. Then substitute [ $H_{3}O^{+}$ ] into the pH expression to find the pH.	KOH(aq) → $K^{+}(aq) + OH^{-}(aq)$ [OH $^{-}$ ] = 0.225 M $[H_{3}O^{+}][OH^{-}] = K_{w} = 1.00 \times 10^{-14}$ [ $H_{3}O^{+}$ ](0.225) = 1.00 × 10 $^{-14}$ [ $H_{3}O^{+}$ ] = 4.44 × 10 $^{-14}$ M pH = -log[ $H_{3}O^{+}$ ] = -log(4.44 × 10 $^{-14}$ ) = 13.353
(b) Since Sr(OH) $_{2}$ is a strong base, it completely dissociates into 1 mol of Sr $^{2+}$ and 2 mol of OH $^{-}$ in solution. The concentration of OH $^{-}$ is therefore, twice the given concentration of Sr(OH) $_{2}$ . Use this concentration and K $_{w}$ to find [ $H_{3}O^{+}$ ]. Substitute [ $H_{3}O^{+}$ ] into the pH expression to find the pH.	$Sr^{2+}(aq) → Sr^{2+}(aq) + 2 OH^{-}(aq)$ [OH $^{-}$ ] = 2(0.0015) M = 0.0030 M $[H_{3}O^{+}][OH^{-}] = K_{w} = 1.0 \times 10^{-14}$ [ $H_{3}O^{+}$ ](0.0030) = 1.0 × 10 $^{-14}$ $[H_{3}O^{+}] = 3.3 \times 10^{-12}$ M pH = -log[ $H_{3}O^{+}$ ] = -log(3.3 × 10 $^{-12}$ ) = 11.48

Question 15.10: Finding the [OH-] and pH of a Strong Base Solution What is t...

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