Question

A volume of 500.0 mL of 0.150 M NaOH is added to 525 mL of 0.250 M weak acid (?a=4.43×10−5). What is the pH of the resulting buffer? HA(aq)+OH−(aq)⟶H2O(l)+A−(aq) pH=

Answer 1

NaOH 100 No of moles of 500 mL of 0.150 m NaOH = 500 x x 0.150 mol/x = 0.075 mol No. of moles of HA in 525 ml of 0.250 M of HA = 525 x x 0.250 mol/x 1000 = 0.13 mol Now according to the given so eac. tion 1 mol HA reacts with 1 mol OH to give 1 mol A- (salt) Thus - HA (ar) + oH (ar) Hao (1) + A (ar) 1 mol 1 mol 1 mol 0.075 mol 0.075 mol 0.075 mol No. of moles of unreacted HA in the solution = (0.13 – 0.075 mol = 0.055 mol

: No. of moles of Á in the solution = 0.075 mol Total volume of the solution = (500 + 525) mL = 1025 ml = 1.025 L . Molarsity of HA 0.055 mol 1-025 L = 0.054 mol/L Molarsity of A 0.075 mol [A-] = 1.0251 = 0.073 mol/L Given ka=4.48x10-5 a - log ka = -log (4.43 x 10-5) a pka = 4.35

Now, according to Henderson equ- ation, the pH of that buffer can be given as - 1 pH = pka + log [A-] [HA] given = 4.35 + log 0.073 0.054 = 4.35 + 0.13 - 4.48 Thus the pH of the roesulting buffer solution .) = 4.48 Aus