Answer
a. The major species present in that solution are: $K^{+}, OH^-, H_2O$.
$[OH^-] = 0.015M$
$pH = 12.18$
b. The major species present in that solution are: $Ba^{2+}, OH^-, H_2O$
$[OH^-] = 0.030M$
$pH = 12.48$
Work Step by Step
a.
Potassium hydroxide reacts on aqueous solution, following the equation:
$KOH(aq) -- \gt K^+(aq) +OH^-(aq)$
Since it is a strong base, it will react completely, so, there is no considerable $KOH$ at the equilibrium.
So, the major species present in that solution are: $K^+, OH^-, H_2O$ .
- Since $KOH$ reacts completely : $[OH^-] = [KOH]_{initial} = 0.015M$
1. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 0.015)$
$pOH = 1.82$
2. Find the pH:
$pH + pOH = 14$
$pH + 1.82 = 14$
$pH = 12.18$
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b.
Barium hydroxide reacts on aqueous solution, following the equation:
$Ba(OH)_2(aq) -- \gt Ba^{2+}(aq) +2OH^-(aq)$
Since it is a strong base, it will react completely, so, there is no considerable $Ba(OH)_2$ at the equilibrium.
So, the major species present in that solution are: $Ba^{2+}, OH^-, H_2O$ .
- Since $KOH$ reacts completely, and each $Ba(OH)_2$ produces 2 $OH^-$
$0.015M(Ba(OH)_2) \times \frac{2mol(OH^-)}{1mol(Ba(OH)_2)} = 0.030M (OH^-)$
1. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 0.03)$
$pOH = 1.52$
2. Find the pH:
$pH + pOH = 14$
$pH + 1.52 = 14$
$pH = 12.48$
