Question

You prepared a phosphate buffer solution by mixing 40.0 mL of 0.250 M KH_2PO_4 and 60.0 mL ol 0.450 M K_2HPO_4 solution. (a) What are the molar concentrations of H_2PO_4 and HPO_4^2-, respectively, in the final buffer solution? (b) Calculate the pH of the solution. (c) Write a net ionic equation when a strong acid, such as HCl(aq) is added to the phosphate buffer solution. (d) Calculate the final pH when 0.0060 mol of HCl is added to the buffer solution? (Assume the volume of the solution remains at 100.0 mL.) (The of H_2PO_4 is K_a = 6.2 x 10^-8)

Answer 1

40 mL of 0.25 M KH2PO4 and 60 mL of 0.45 M K2HPO4

total volume is 100 mL which contains

40 mL x 0.25 moles/L of KH2PO4 = (40 x 0.25)/1000 moles = 0.01 moles in 100 mL = (0.01 x 1000)/100 = 0.1 M KH2PO4 in the final solution

60 mL x 0.45 moles/L of K2HPO4 = (60 x 0.45)/1000 moles = 0.027 moles in 100 mL = (0.027 x 1000)/100 = 0.27 M K2HPO4 in the final solution

Use the Henderson Hasselbalch equation

pH = pKa + log [A- ] / [HA]

pH = 7.20 + log (0.27) / (0.1)

pH = 7.20 + log 2.7

pH = 7.63

HCl will react with HPO₄^2- to produce H₂PO₄^-

net ionic equation H⁺(aq) + HPO₄^2-(aq) $\rightarrow$ H₂PO₄^-(aq)

The added HCl will react with K2HPO4 as shown above

so the concentration of K2HPO4 will decrease and KH2PO4 will increase

the new concentration of K2HPO4 will be 0.027 moles - 0.0060 mol = 0.021 mole in 100 mL = 0.21 M

The new concentration of KH2PO4 will be 0.01 moles + 0.0060 mol = 0.016 mole in 100 mL = 0.16M

The pH will be

pH = pKa + log [A- ] / [HA]

pH = 7.20 + log (0.21) / (0.16)

pH = 7.20 + log 1.31

pH = 7.32