Question

A buffer is prepared by mixing 122 mL of a 0.120 M solution of the weak acid HA with 145 mL of a 0.150 M solution of NaA. What will be the pH of the resulting solution?  K_a for HA is 2.30x10^-6. all volumes should have 3 decimal places

Answer 1

Ans---

First we have to calculate initial number of moles of acid and base present in the buffer solution

number of moles of HA=0.120*.122L=0.014 mole

number of moles of NaA=0.150*0.145L=0.021mole

HA is dissociates to form conjugated acid and base as follows

HA <.......> H⁺ + A^-

^{initial concentration 0.014 0 0.021}

^{at equilibrium o.o14-X x 0.021+x}

   ^{Ka=[H+][A-]/[HA]}

^{2.30*10^-6=X(0.021+X)/(0.014-X)}

^{since the solution is very dilute so +x,-x is ignored}

0.021x=0.014*2.30*10^-6

x=1.523*10^-6=[H+]

pH=-log[H+]

pH=5.81