Question

A: 48.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 24.0 mL of KOH at 25 ∘C

B: 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.

C: 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

Answer 1

Part.1

Given reaction is: HBr+ KOHKBr H2O Molarratio between HBr and KOH is 1 :1 So, mmol of HBr = 48.0 mL x 0.25 M = 12 mmol of KOH = 24.0 mL x 0.50 M = 12 Since, mmol of both the acid and base are same, Therefore pH of the resulting solution will be 7.
Part.2

Net ionic equation for the reaction of NH3 and HNO3 is as follows One mole of NHa reacts with one mole of HNO3. So, mole ratio between NH3 and HNO3 is mmol NH, = 50.0 mL × 0.20 M = 10 mmol mmol HNO, - 50.0 mL x0.20 M - 10 mmol Therefore, after the mixing neither NH3 nor HNO3 remains. Solution contains only NH^ and NO,. Nitrate ion not hydrolyzed but ammonium ion is hydrolyzed. NHt1- = 0.10M (50.0 + 50.0) mL Set ICE table: Initial (M): Change(M): -x Equilibrium (M): 0.10-x 0.10 1.0x 10-14 1.8x10" (0.10-x) 5.6x10-10- Solving we get .r - 7.5x10-* (0.10-x so, 1 H,0. | = 7.5x10" M

so, LH,O] = 7.5x10-6 M pH =-log[H,0.] -log(7.5x 10) = 5.12

Part.3 :

Net ionic equation for the reaction of CH,COOH and NaOH is as follows: CH.COOH+OH' → CH.COO-+H.O One mole of CH,COOH reacts with one mole of NaOH So, mole ratio between CH,COOH and NaOH is 1:1. mm01 CH3COOH = 30.0 mL × 0.50 M-15 mmol mmol NaOH = 30.0 mL x 0.50 M = 15 mmol Therefore, after the mixing neither CH COOH nor NaoH remains. CH.COO" + H,0 F-CH.COOH + ОН 15 mmol CH COO 0.25 MM (30.0 + 30.0) mL Set ICE table: CH.COO-+H.。근 CH.COOH + OH Initial(M):0.25 Change(M):n Equilibrium (M): 0.25-x CH.COO 1.0x 10-14 1.8 x10 5.6 × 10:10 Solving we et1.2x10 0.25 -.r (0.25-x)

= 1.2×10-5 M So,1 OH pOH =-log[OH'--log(1.2x10") = 4.92 pH = 14.00-p0H 14.00-4.92 = 9.08