Question

Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added to (a) 33.0 mL of 0.230 M NaOH(aq). pH = (b) 13.0 mL of 0.330 M NaOH(aq). pH =

Answer 1

Calculate the pH of the resultng solution if 23.0 mL of 0.230 M HCI(aq) is added to (a) 33.0 mL of 0.230 M NaOH(aq). Number pH- 12.61 (b) 13.0 mL of 0.330 M NaOH(aq). Number pH- moles of HCl = (0.230 M) x (0.023 L) = 0.00529 mol moles of NaOH (0.230 M) x (0.033 L) 0.00759 mol Here; NaOH is excess. So solution is basic excess NaOH = [OH-] = (0.00759-0.00529) / (0.056 L) = 0.0410 M pOH = 1.39 pH 14-pOH 12.61 moles of HCl = (0.230 M) x (0.023 L) = 0.00529 mol moles of NaOH (0.330 M) x (0.013 L) 0.00429 mol Here; HCl is excess. So solution is acidic excess HCI-H+] = (0.00529-0.00429)/ (0.036 L,-0.0278 0.0410 M pH-1.56