Answer to Question #117700 in Chemistry for Yehya Mohamed

Question

Calculate the pH of the solution that was formed after mixing of 150 mL of 0.200 M H3PO4 with 150 mL of 0.266 M NaOH. Remember to use the Henderson-Hassselbach equation and appropriate pKa value choosing one from pK1=2.12, pK2=7.21, pK3=12.32

Answer 1

When 150 mL of 0.200 M H₃PO₄ reacts with 150 mL of 0.266 M NaOH, H₃PO₄ is converted into NaH₂PO₄ (0.200-0.066 = 0.134 M) and into Na₂HPO₄(0.066 M). Therefore, a buffer solution is formed, the buffering action of which is based on the following reaction of dissociation of H₂PO₄^- (conjugate acid) into HPO₄^2- (conjugate base) and H⁺:

H₂PO₄^- $\rightarrow$ HPO₄^2- + H⁺.

The appropriate pK_a value is then pK_a2=7.21. According to the Henderson-Hasselbach equation, the pH of a buffer solution is:

$pH = pK_a + \text{log}\frac{[HPO_4^{2-}]}{[H_2PO_4^-]} = 7.21 + \text{log}\frac{0.066}{0.134} = 6.90$

Answer: the pH of the solution is 6.90.