Calculate the pH of a solution prepared by dissolving 1.2 g of potassium hydroxide (KOH) in 1,250 mL of water.
A)0.017
B)1.77
C)9.22
D)12.23

Question

Quizplus · Accepted Answer

KOH is a strong base that completely dissociates in water. To find the pH, first calculate the molarity of the KOH solution. The molar mass of KOH is approximately 56.11 g/mol. Thus, 1.2 g of KOH is equivalent to $1.2 \, \text{g} \div 56.11 \, \text{g/mol} = 0.0214 \, \text{mol}$. The volume of the solution is 1,250 mL or 1.25 L. Therefore, the concentration of KOH (which equals the concentration of $OH^-$ ions due to complete dissociation) is $0.0214 \, \text{mol} \div 1.25 \, \text{L} = 0.0171 \, \text{M}$.The $pOH$ can be calculated using the formula $pOH = -\log[OH^-]$. Substituting the concentration of $OH^-$ gives $pOH = -\log(0.0171) = 1.77$.Since $pH + pOH = 14$ for aqueous solutions at 25°C, the $pH$ is $14 - 1.77 = 12.23$.

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