Finding pH from new solution

Question

1.00L buffer made with 1.02 M generic acid HA and its conjugate salt 0.9 M NaA has a pKa of 4.87. If you add 51.0 mL (0.051 L) of 0.456 M HNO3, what is the new pH?

Answer 1

The addition of 51.0 ml if HNO3 results in the addition of 0.456/1000 x 51.0 = 0.0233 moles of HNO3. These number of moles reacts with the NaA salt to produce 0.9 - 0.0233 = 0.877 moles of NaA and 1.02 + 0.0233 = 1.043 moles of HA. As the new volume is 1051 mL, the new concentrations of NaA and HA are 0.834 M/L and 0.992 M/L respectively. (eg. the new concentration of HA is obtained by 1.043 mol/1051mL x 1000 mL/L = 0.992M)

Put these new concentrations into the buffer equation pH = pK_a + log(C_s/C_a)

pH = 4.87 +log(0.834/0.992) = 4.87 + log 0.8 = 4.87 + log( 8 x 10^-1) = 4.77