Question

Determine the pH and pOH of 0.25 L of a solution that is 0.0191 M boric acid and 0.0286 M sodium borate; pKa for B(OH)3 = 9.0 at 25°C.

Answer 1

The pH of the solution is approximately 9.176, and the pOH is approximately 4.824.

Step-by-step explanation:

To determine the pH and pOH of the solution, we need to calculate the concentrations of the H+ and OH- ions.

Boric acid (H3BO3) and sodium borate (Na2B4O7) are a conjugate acid-base pair, so we can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log([base]/[acid])

Given that the pKa for B(OH)3 is 9.0, [base] = 0.0286 M and [acid] = 0.0191 M:

pH = 9.0 + log(0.0286/0.0191) = 9.0 + log(1.496) = 9.0 + 0.176 = 9.176

So, the pH of the solution is approximately 9.176.

To calculate the pOH, we can use the following relationship:

pH + pOH = 14

Therefore, the pOH of the solution is approximately 4.824.