Question

A volume of 500.0 mL of 0.160 M NaOH is added to 575 mL of 0.200 M weak acid (K, the resulting buffer? 5.93 x 10). What is the pH of HA (aq) +OH (aq)HO) + A (aq) pH =

Answer 1

Number of moles of weak acid = mass/mol.wt. = M*V = 0.200 M * 575 mL = 115 mmol Number of moles of NaOH = M*V = = 0.160 M * 500 mL = 80 mmol When weak acid reacts with NaOH, its conjugate base is formed. HA + NaOH + H2O Initial 115 Add - 80 Change -80 Final Final 35 35 0 80 The resulting solution contains only weak acid and its conjugate base. So, this is a buffer solution. pH of the buffer solution is determined by Henderson equation, pH = pKa+log [base] [acid] 80 35 pH =-log(5.93x10^') + logo pH = 4.2255+0.359022 = 4.5845 pH = 4.58 pH = 4.58

The balanced equation is NaOH(aq) + HCl (aq) -----------> NaCl (aq) + H20 (g) Number of moles of HCl = 0.350 M * 35.0 mL = 12.25 mmol Number of moles of NaOH = 0.350 M * 45.0 mL = 15.75 mmol Net moles of NaOH = 15.75 – 12.25 = 3.5 mmol Total volume = 35.0 +45.0 = 80 mL Conc. of NaOH = [OH-] = 3.5 mmol/80.0 mL = 0.04375 M pOH = -log[OH-] = -log(0.04375) = 1.36 pH = 14 - pOH = 14 – 1.36 = 12.64 The balanced equation is NaOH(aq) + HCl (aq) -----------> Naci (aq) + H20 (g) Number of moles of HCl = 0.350 M * 35.0 mL = 12.25 mmol Number of moles of NaOH = 0.450 M * 25.0 mL = 11.25 mmol Net moles of HCl = 12.25 – 11.25 = 1.0 mmol Total volume = 35.0 +25.0 = 60 mL Conc. of HCI = [H+] = 1.0 mmol/60.0 mL = 0.016667 M pH = -log[H+] = -log(0.016667) = 1.78