Calculate the pH of the resulting solution if 22.0 mL of 0.220 M HCl(aq) is added to (a) 32.0 mL of 0.220 M NaOH(aq). (b) 12.0 mL of 0.320 M NaOH(aq). Solution number of moles of HCl = 22*0.22 = 0.00484 mol number of moles of NaOH = 32*0.22 = 0.00704 mol number of moles excess = 0.00704 - 0.00484 = 0.0022 pOH = -log([OH-]) pOH = -log(0.0022*1000/(22+32)) pOH = 1.39 pH = 14 - 1.39 = 12.61 number of moles of NaOH = 12*0.32 = 0.00384 mol number of moles excess = 0.00484 - 0.00384 = 0.001 mol pH = -log([H+]) pH = -log(0.001*1000/(12+22)) pH = 1.531 .