Question

Calculate the pH of the solution when the following substances are added together: ​

20 mL of 0.001 M HCl and 0.5 mL of 0.04 M Sodium Acetate

200 mL of 0.1 M HCl and 8 mL of 2.5 M Sodium Acetate

Answer 1

1)

millimoles of CH3COO- = 0.5 x 0.04 = 0.02

millimoles of HCl = 20 x 0.001 = 0.02

CH3COO-   + HCl   -----------------> CH3COOH

    0.02            0.02                              0

    0                    0                             0.02

concentration of CH3COOH = 0.02 / (20 + 0.5) = 9.756 x 10^-4 M

Ka = x^2 / 9.756 x 10^-4 - x

1.8 x 10^-5 = x^2 / 9.756 x 10^-4 - x

x = 1.238 x 10^-4

[H+] = 1.238 x 10^-4 M

pH = -log (1.238 x 10^-4 )

pH = 3.91

similarly do 2nd one also.