Question

Please explain to me how I solve this

Calculate the pH of a 0.2 M acetate buffer (pKa = 4.77) that contains twice as much acid as conjugate base. Calculate the pH of the solution that results following the addition of 10mL of 1M NaOH to 40mL of 1M NH_4C1 Ka = 5.62 times 10^-10

Answer 1

Acc^n to Henderson - Hasselbaleh equation P^H = P^Ka + log_10 [Salt]/[acid] As [acetic acid] = 2[sod . Acetate] [0.2m] = [0.4m] P^H = 4.77 + log_10 0.4/0.2 4.77 + 0.3010 5.07 P^H = 5.07 \r\n Moles of NaOH added = Molarity times Volume/1000 For 10 ml of 1 M NaOH (x) no of moles = 1 times 10/1000 = 0.01 For 40ml of 1 M NH_4Cl n = 1 times 40/1000 = 0.04 given K_a = 5.62 times 10^-10 we know P^Ka = -log_10^Ka Therefore P^Ka = -log_10(5.62 times 10^-10) - [log_10 5.62 + log 10^-10] - [0.75 - 10] 9.25 we know P^Ka + P^Kb = 14 P^Kb = 14 - 9.25 = 4.75 Acc^n to Henderson's equation P^OH = P^Kb + log_10 [salt]/[base] P^OH = 4.75 + log_10 0.04/0.01 4.75 + 0.602 P^H + P^OH = 14 P^H = 14 - P^OH = > 14 - 5.35 = 8.65 P^H = 8.65