Questions: Calculating the pH of a strong base solution 0 / 5 Ava A chemist dissolves 586. mg of pure barium hydroxide in enough water to make up 130 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 find the number of moles of barium hydroxide (\(\text{Ba(OH)}_2\)) dissolved. The molar mass of \(\text{Ba(OH)}_2\) is calculated as follows: \[ \text{Molar mass of Ba(OH)}_2 = 137.33 \, (\text{Ba}) + 2 \times 16.00 \, (\text{O}) + 2 \times 1.01 \, (\text{H}) = 171.35 \, \text{g/mol} \]

Given mass of \(\text{Ba(OH)}_2\) is 586 mg, which is 0.586 g. The number of moles is: \[ \text{Moles of Ba(OH)}_2 = \frac{0.586 \, \text{g}}{171.35 \, \text{g/mol}} = 0.003419 \, \text{mol} \]

Next, we calculate the concentration of \(\text{Ba(OH)}_2\) in the solution. The volume of the solution is 130 mL, which is 0.130 L. The concentration is: \[ \text{Concentration of Ba(OH)}_2 = \frac{0.003419 \, \text{mol}}{0.130 \, \text{L}} = 0.02630 \, \text{M} \]

Barium hydroxide dissociates completely in water: \[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^- \]

Thus, the concentration of \(\text{OH}^-\) ions is twice the concentration of \(\text{Ba(OH)}_2\): \[ [\text{OH}^-] = 2 \times 0.02630 \, \text{M} = 0.05260 \, \text{M} \]

The pOH is calculated using the concentration of hydroxide ions: \[ \text{pOH} = -\log_{10} [\text{OH}^-] = -\log_{10} (0.05260) = 1.279 \]

Finally, we use the relationship between pH and pOH: \[ \text{pH} = 14 - \text{pOH} = 14 - 1.279 = 12.721 \]

Final Answer