Question

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 12.0 mL of KOH.

Express your answer numerically.

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 13.0 mL of HNO3.

Express your answer numerically.

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH.

Express your answer numerically.

Answer 1

HBr molarity = 8-15m | KoH: Molarity = 0.25M volume = soml= 0.052 volume = 12 m) = 0.0122 no of moles=0.85X0.012 no of moles 0.15 X0-05 0.003 0.0075 M2 2 HBr + KOH - KBr + to 0.0045 0.07258M Io.005 0.003 -0.003 -0.003 -0.004510 Total volume 0-054 0-012 = 0.062L [1739]>Lit+]=> 0-062 pH = -log[HT] pH = -log (0-07258) = 1-139 ANO: Molarity = 0.5M NH = Molarity = 0.2M I volumea 13ml=0.0132 I no of moles o 5XO-023 = 0.0065 nouit moles = 0.24 0.075 = 0.015 NH & ANG NHA NO = 1.8x105 volume = 7.5m= 0.075L pkb = -logo an10g (1-84105) pkb _ 4.745 0.015 10.0065 lo -0.0065 --0065 +0.0065 Elo.0085 o 10.0065 Tota) volume 20.075+0.013 = 0.0882 ENH₂] = 33 V0-018 Lolta preblog [INTANG 0.0085 20.09659M. Inity 0.0065 0.07386M [NH Ng] = 0.088 0.07386 - 4-245 polt= pkb log 0.09659 + 109 0.07386 Jo.0259 --4-63 9.37 pH = 14-polt = 14-4.632

CH coolt: molarity 0.35m volume =52m)= 0.0522 Nast: molarity = 0.4m volume=33ml = 0.0332 no ob moles= 0.4 X0-033 = 0.0132 m 2 no of moles a 0.35X0.052 ao.0182 et coolt & Nash - CH Coona tho 10-0182 0.0132 0. C -0.0132 -0-0 132 to-0132 Leloucos o l0.013 2 1 Toral volume = 0.052+0 -033 = 0.0852 0.005 [CH, Cool+3 = 3 2 0-085 -20.85882 M 2CH COONara 0.0132 = 20.15529 M 3 0 -085 2 kaa 1-88105 plea = -log ka a-log (1-88105) 24-745 {CH Coona] Hapka tlog {ch coolt} 0.15529 = 5.167 pH = 4.745 +log 0.05882