Calculate the mass of sodium acetate that must be added to 100 ml of a 0.15 M solution of chlorhydric acid to prepare a buffer with a pH value of 5.O? (Ka= 1.8 x 10^-5)

Question

I am looking for the steps showing how to solve this problem. Thank you! :)

Answer 1

If we let x₀ be the molarity of the NaAc (sodium Acetate) after adding the solid to 100 ml of water. We expect that the Ac^- will react with the H⁺ of the HCl until a pH of 5 is reached. The equation for the equilibrium of the conjugate acid HAc is

HAc → H⁺ + Ac^- which has a K_A = 1.8 x 10^-8

We can write the K expression in terms of x the molarity of Ac^- that forms Acetic acid in order to reach equilibrium:

K_A = [H⁺][Ac^-]/[HAc] = [.15 - x][x₀-x]/x

Note that .15-x = 10^-5 (pH of 5) and x = .15 - 10^-5

1.8 x 10^-5 = 10^-5 *(x₀-x)/x or x₀ = 2.8x which is approximately 2.8*.15 (sing 10^-5<<.15) =0 .42 molar

If you want to make a .42 molar NaAc solution, you need 0.042 moles for 100 ml (one tenth of a liter, one tenth the moles)

Another way to do this is to use the Henderson-Hasselbalch equation with the understanding that the acetate will neutralize the HCl and that .15M HAc is formed:

pH = pK_A + log([Ac^-]/[HAc]) or the log term becomes log [x₀-.15]/[.15 If you solve this you get the same value for x₀