Question

2. Calculate the percent change in pH upon addition of 0.010 mol of NaOH to 1.0L solution of following two systems: 20.50 mol HA and 0.50 mol A., initial pH= 5.00 6rpotpr buffer Copot b) 0.050 mol HA and 0.050 mol A-, initial pH=5.00

Answer 1

Given that Initial pH = 5.00 [base] pH = pka +log [acid] 0.50 5.00 = pKa+logo50 5.00= pka+0 pKa = 5.00

(a) Number of moles of NaOH added = 0.010 mol When NaOH adds to the buffer solution, it reacts with HA to form A. So the conc. of A-increases and conc. of HA decreases. acid base + - E + H₂O HA 0.50 0 0.50 Initial Add OH 0 0.010 0.50 Change -0.010 -0.010 +0.010 0.51 Equilibrium 0.49 0.51 [base] pH = pka+log Tacid] 0.51 pH = 5.00+log = pH = 5.00+0.017374 = 5.017374 pH change = 5.017374 – 5.00 = 0.017374 = 0.0174 percent change in pH = 1:0114x100 = 0.35% 5.00

(b) Number of moles of NaOH added = 0.010 mol When NaOH adds to the buffer solution, it reacts with HA to form A. So the conc. of A increases and conc. of HA decreases. acid base HA + OH - A + H₂O Initial 0.050 0.050 Add 0 0.010 Change -0.010 -0.010 + 0.010 Equilibrium 0.04 0.06 [base] pH = pka+log [acid] 0.06 pH = 5.00+log = pH = 5.00+ 0.1761 = 5.1761 pH change = 5.1761-5.00 = 0.1761 , 0.1761 percent change in pH = x 100 = 3.52% 5.00