Question

A solution has 0.100 M HC7H5O2 and 0.200 M Ca(C7H5O2)2. The solution volume is 5.00 L. What is the pH of the solution after 2.00 g NaOH is added? The Ka for HC7H5O2 is 6.3 × 10-5. If 0.02 moles of HCl was added to the solution described above, what change do you expect to occur in the pH?

Answer 1

From Henderson equation, pH of a buffer solution is given by

pH = pKa+log[salt]/[acid]

Given that, Ka = 6.3*10^-5, pKa = -logKa = 4.20

[salt]=[ca(C7H5O2)2] = 0.2moles/5 L = 0.04 M

[acid] = [HC7H5O2] = 0.1moles/5 L = 0.02 M

pH = 4.2+log(0.04/0.02) = 4.50

If 2.0 g of NaOH is added i.e., 2/40 = 0.05/5 =0.01M, it reacts with acid and forms salt and water. Thus, the concentration of acid decreases by 0.01 M and concentration of salt increases by 0.01 M.

Now, pH of solution is given by

pH = 4.20+log((0.04+0.01)/(0.02-0.01)) = 4.89

If 0.02 moles of HCl is added to the above solution, it reacts with salt and decreases its concentration and forms acid. Hence the concentration of acid increases by 0.02/5 = 0.004 M

Now, pH = 4.20+log((0.05-0.004)/(0.01+0.004)) = 4.71