(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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Chapter16: Acid-base Equilibria
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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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(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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