Questions: A chemist dissolves 731 mg of pure potassium hydroide in enough water to make up 120 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25°C.) Round your answer to 3 significant decimal places.

Solution Steps

First, we need to calculate the number of moles of potassium hydroxide (KOH) dissolved in the solution. The molar mass of KOH is approximately:

\[ \text{Molar mass of KOH} = 39.10 \, (\text{K}) + 15.999 \, (\text{O}) + 1.008 \, (\text{H}) = 56.107 \, \text{g/mol} \]

Convert the mass of KOH from milligrams to grams:

\[ 731 \, \text{mg} = 0.731 \, \text{g} \]

Calculate the moles of KOH:

\[ \text{Moles of KOH} = \frac{0.731 \, \text{g}}{56.107 \, \text{g/mol}} = 0.01302 \, \text{mol} \]

The molarity (M) of the solution is the number of moles of solute per liter of solution. The volume of the solution is given as 120 mL, which is 0.120 L.

\[ \text{Molarity (M)} = \frac{0.01302 \, \text{mol}}{0.120 \, \text{L}} = 0.1085 \, \text{M} \]

Potassium hydroxide (KOH) is a strong base and dissociates completely in water:

\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \]

The concentration of hydroxide ions [OH\(^-\)] is equal to the molarity of the KOH solution:

\[ [\text{OH}^-] = 0.1085 \, \text{M} \]

The pOH of the solution is calculated using the formula:

\[ \text{pOH} = -\log_{10}([\text{OH}^-]) = -\log_{10}(0.1085) = 0.964 \]

The pH is related to the pOH by the equation:

\[ \text{pH} = 14 - \text{pOH} = 14 - 0.964 = 13.036 \]

Final Answer