(a) A 20.00 mL solution of 1.50 M methylamine, CH3NH2 is titrated with 1.50 M solution of hydrochloric acid, HCl. Equation is as given below. [Kb for CH3NH2 = 4.38 × 10-4] CH3NH2(aq) + HCl(aq) → CH3NH3Cl(aq) (i) Calculate the pH of the solution when 12.00 mL of HCl is added. (ii) Identify whether the solution in (i) is before, at or after equivalence point. (iii) Identify whether the salt solution at equivalence point is acidic, neutral or basic. (b) In a titration of CH3COOH with aqueous NaOH solution, 6.50 × 10-3 mole of CH3COONa is formed at equivalence point. If the total volume at equivalence point is 60.0 mL, calculate the pH of the solution. [Ka of CH3COOH is 1.79 × 10-5 and Kb of CH3COOis 5.59 × 10-10]
(a) A 20.00 mL solution of 1.50 M methylamine, CH3NH2 is titrated with 1.50 M solution of hydrochloric acid, HCl. Equation is as given below. [Kb for CH3NH2 = 4.38 × 10-4] CH3NH2(aq) + HCl(aq) → CH3NH3Cl(aq) (i) Calculate the pH of the solution when 12.00 mL of HCl is added. (ii) Identify whether the solution in (i) is before, at or after equivalence point. (iii) Identify whether the salt solution at equivalence point is acidic, neutral or basic. (b) In a titration of CH3COOH with aqueous NaOH solution, 6.50 × 10-3 mole of CH3COONa is formed at equivalence point. If the total volume at equivalence point is 60.0 mL, calculate the pH of the solution. [Ka of CH3COOH is 1.79 × 10-5 and Kb of CH3COOis 5.59 × 10-10]
General Chemistry - Standalone book (MindTap Course List)
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.123QP: A 25.0-mL sample of hydroxylamine is titrated to the equivalence point with 35.8 mL of 0.150 M HCl....
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Question
(a) A 20.00 mL solution of 1.50 M methylamine, CH3NH2 is titrated with 1.50 M solution
of hydrochloric acid, HCl. Equation is as given below. [Kb for CH3NH2 = 4.38 × 10-4]
CH3NH2(aq) + HCl(aq) → CH3NH3Cl(aq)
(i) Calculate the pH of the solution when 12.00 mL of HCl is added.
(ii) Identify whether the solution in (i) is before, at or after equivalence point.
(iii) Identify whether the salt solution at equivalence point is acidic, neutral or basic.
(b) In a titration of CH3COOH with aqueous NaOH solution, 6.50 × 10-3 mole of
CH3COONa is formed at equivalence point. If the total volume at equivalence point is
60.0 mL, calculate the pH of the solution.
[Ka of CH3COOH is 1.79 × 10-5 and Kb of CH3COOis 5.59 × 10-10]
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