Question

Using a 0.25 M phosphate buffer with a pH of 7.2, you add 0.75 mL of 0.55 M HCl to 57 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Using a 0.25 M phosphate buffer with a pH of 7.2, you add 0.75 mL of 0.55 M NaOH to 57 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Answer 1

The Ka values for the phosphoric acid are The phosphoric acid has the following ionizations H3PO4 = H2PO4 + H+ pka2 = 2.15 H2PO4 = HPO4-2 + H+ pka = 7.21 HPO4-2 = PO4 -3 + H+ pka, = 12.3 Since the given pH (7.2) is nearer to second ionization. The phosphate buffer must be prepared by H2PO4-1 and HP04-2 [base] pH = pK, +log [acid] => 7.2= 7.21+ log HP01 [HPO,²1 -0.01 log 10% [H,PO, ] [HPO4'] -10-0.01 0.977237 [H2PO4) [HPO, ?]=0.977237x[H,PO, 1 Given that conc. of buffer = 0.25M That is [HPO, ?]+[H PO, ]=0.25 0.977237x[H,PO, -]+[H,PO, ]=0.25 1.977237x[H,PO]=0.25 0.25 [H PO, ]=1077027 = 0.126439 1.977237 [HPO,"]=0.977237x [H,PO]=0.977237x0.126439= 0.123561 [H,PO,"]=0.126439M [HPO, ?]=0.123561M Volume of the buffer solution = 57 mL Number of moles of H,PO, present in the solution = M*V = 0.126439M x 57 mL = 7.207026 mmol Number of moles of HPO-present in the solution = M*V = 0.123561M x 57 mL = 7.042974 mol

(a) Number of moles of HCl added = 0.55 M * 0.75 mL = 0.4125 mmol When HCl adds to the buffer solution, it reacts with HPO-2 to form H,PO.. So the conc. of HPO2-2 decreases and conc. of H PO, increases. base acid HPO4 ? + H+ = H2PO4 Initial 7.042974 7.207026 Add 0.4125 Change -0.4125 - 0.4125 + 0.4125 Equilibrium 6.630474 7.619526 pH = pKa+log base (acid] =pH = 7.21+log [HPO, ?) [H,PO] 6.630474 pH = 7.21+log - = 7.21-0.06038 = 7.149617 = 7.15 8 7.619526 New pH of the reaction solution = 7.15

(b) Number of moles of NaOH added = M*V = 0.55 M * 0.75 mL = 0.4125 mmol When NaOH adds to the buffer solution, it reacts with H,PO, to form HPO-2. So the conc. of HPO2-2 increases and conc. of H2PO4 decreases. acid base H2PO4 + OH = HPO, -2 Initial 7.207026 7.042974 Add 0.4125 Change -0.4125 -0.4125 + 0.4125 Equilibrium 6.794526 7.455474 pH = pKa +log [base] pH = pka+10% (acid] =pH = 7.21+log [HPO, ?) [H,PO, -] 7.455474 =pH = 7.21+10% 6.794526 = 7.21+0.040316= 7.250316 = 7.25 New pH of the reaction solution by addition of NaOH = 7