Question

A 50.0 mL solution of 0.139 M KOH is titrated with 0.278 M HCl. Calculate the pH of the solution after the addition of the following amounts of HCl.

Answer 1

Number of mole of solution calculate by its Molarity x volume Total volume is 50 mL 0.05L a) 0.00 mL HC1 Moles of KOH = Molarity x volume 0.139% 0.05 L = 6.95x10-3 mol Moles of HC Molarity xvolume = 0.278×0.00し = 0.0mol 0 0.00695 TOH] 0.050.139 Then pOHlog(OH') =-log(0.139) = 0.86 pH- 14 pOH 14-0.86- 13.14 b) 6 mL HC Moles of KOH= Molarity×volume = 0.1 39% 0.05 L = 6.95 x10" Moles of HC oty xvolume = 0.278× 0.006し 0.001668mol

0.001668 mol of HC1 neutralized. The remaining KOH solution calculated 0.00695-0.001668 0.005382 mol as Total volume 50+6-56 mL = 0.056L [OH-1-0.005382 0.056 0.096 pOHlog(OH) =-log(0.096) = 1.02 pH-14 -pOH- 14-1.02- 12.98 C) 12.5 mL HCl Moles ofKOH= Molarity ×volume -0.139% 0.05 L 6.95 x10" Molarity× volume = 0.278× 00125L Moles ofHCI 0.003475 mol 0.003475 mol of HCl neutralized. The remaining KOH solution calculated 0.00695-0.003475 Total volume 50+12.5 62.5 mL = 0.0625し 0.003475mol [OH-I-0.003475 0.06250.056 pOHlog(OH) log(0.056) = 1.252 pH = 14-pOH = 14-1252 = 12.74 d) 19.0 mL HCl

Moles ofKOH= Molarity x volum e = 0.139% 0.05 L 6.95 x10 Moles of HC Molarity x volume = 0.278× 0.019L 0.005282 mol 0.00695-0.005282 0.001668 mol 0.005282 mol of HCl neutralized. The remaining KOH solution calculated Total volume 50+19.0 = 69 mL 0.069L 0.001668 [OH-]=-0069 0.024 pOHlog(OH') =-log(0.024) =1.6197 pH = 14-pOH = 14-1.6197 = 12.38 e) 24.0 mL HC1 Molarity× volume = 0.139% 0.05 L = 6.95 x10 Molarity × volume = 0.278× 0.024L = 0.006672 mol Moles ofKOH= Moles ofHCI 0.006672 mol of HC1 neutralized. The remaining KOH solution calculated0.00695-0.006672 0.000278 mol Total volume = 50+24.0 = 74 mL = 0.074L

0.000278 [OH-] 0,074 pOH -log(OH) = 2.425 0.00376 =-log(0.0376) pH = 14-pOH = 14-2.425-1 1.575 f) 25.0 mL HCI Moles ofKOH= Molarity x volume -0.139% 0.05 L = 6.95 × 10-3 Moles of HC Molrity xvolume = 0.278× 0.025L 0.00695 mol 0.00695 mol of HCl neutralized. The remaining KOH solution calculated0.00695-0.00695 =0.00 mol All KOH will neutralized at equivalence point [H] [OH ] and the pH of the solution is7 g) 26.0 mL HCI Moles ofKOH= Molarity x volume 0.139x 0.05L 6.95 x103 0.278×0.026し 0.007228 mol Moles of HC Molarity x volume 0.007228 mol ofHCl neutralized. The remaining HCl solution calculated = 0.007228-0.00695 0.000278 mol Total volume = 50+26.0 = 76 mL = 0.076L

1-0000278=000366 0.076-= 0.00366 pH =-log(0.00366) pH 2.44 g) 31.0 mL HCl Moles of KOH= Molarity× volume -0.139% 0.05 L = 6.95×10-3 Moles of HC Molity xvolume 0.278x0.031L 0.008618mol 0.008618 mol of HCl neutralized. The remaining HCl solution calculated0.008618-0.00695 -0.001668mol Total volume = 50+31.0 = 81 mL = 0.081L RT= 0.081 0.00166 0.021 pH log(H') pH og(0.021) pH 1.68