Question

Part B

A 96.0 mL volume of 0.25 M  HBr is titrated with 0.50 M  KOH. Calculate the pH after addition of 48.0 mL of KOH at 25 ∘C.

Express the pH numerically.

Part C

Consider the titration of 50.0 mL of 0.20 M  NH3 (Kb=1.8×10−5) with 0.20 M  HNO3. Calculate the pH after addition of 50.0 mL of the titrant at 25 ∘C.

Express the pH numerically.

Part D

A 30.0-mL volume of 0.50 M  CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25 ∘C.

Express the pH numerically.

Answer 1

Part B: Reaction: OH → KBr+H2O One mole HBr reacts with one mole KOH. So, mole ratio between HBr and KOH is 1:1 mmol HBr 96.0 mL 0.25 M = 24 mmol mmol KOH = 48.0 mL x 0.50 M 24 mmol Therefore, after the mixing neither HBr nor KOH remains. Solution contains only KBr and H2O and both are neutral Hence, the pH of the solution is 7.00

Set ICE table Initial (M) Change(M): -r Equilibrium(M): 0.10-x 0.10 tx tx н.o K, NH 1.0× 10-14 1.8×10-5 (0.10-x) 5.6x 10-10 Solving we get x = 7.5x 10" So, | H,0. | = 7.5×10-6 M pH =-log[H,0.」 (0.10-x log(7.5x10") 5.12

Part D: Net ionic equation for the reaction of CH,COOH and NaOH is as follows: One mole of CH,COOH reacts with one mole of NaOH So, mole ratio between CH,COOH and NaOH is 1:1 mrmol CII,COO 30.0 m.x).50M :15 mmol mmol NaOH30.0 mL x 0.50 M- 15 mmol Therefore, after the mixing neither CH,COOH nor NaOH remains. CH.COO-+ H,0 근 CH.COOH + OH. 15 mmol [CH,coo 30.0+300 m 0.25 M

Set ICE table CH.COO. +H.O 근 CH.COOH+ OH 0.25 Initial(M) Change(M):-r Equilibrium(M): 0.25-r K-K,ー[CH.COOHLOH.1 +x +x K, CH COO 1.0x10-4 1.8×10% (0.25-x) 5.6x10-10- Solving we get x = 1.2x10-5 So, | OH- | = 1.2×10° M. pOH=-log OH-」-log(1.2x10") pH 14.00- pOH 0.25-x = 4.92 14.00-4.92 = 9.08