Answer:
The pH of the final solution is 7.15
Explanation:
50 mL of 2.0 M of [tex]K_2HPO_4[/tex] and 25 mL of 2.0 M of [tex]KH_2PO_4[/tex] were mixed to make a solution
Final volume of the solution after dilution = 200 mL
Final concentration of [tex]K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M[/tex]
Final concentration of[tex]KH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M[/tex]
We use Hasselbach- Henderson equation:
[tex]pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85[/tex]
Substituting the values:
[tex]pH = 6.85+ log \frac{0.5}{0.25}pH = 6.85+ log 2pH = 6.85+ 0.3 = 7.15[/tex]
Therfore, the pH of the final solution is 7.15
