Question

Calculate pH of a weak acid/conjugate base buffer solution.

1. a) Calculate the pH of 650. mL of a 0.211-M solution of acetic acid before and after the addition of 0.123 mol of potassium acetate.

pH befor addition =
pH after addition =

1 b) Calculate pH of a weak base/conjugate acid buffer solution.

A 0.190-M aqueous solution of C₂H₅NH₂ (ethylamine) has a pH of 11.9. Calculate the pH of a buffer solution that is 0.190 M in C₂H₅NH₂ and 0.487 M in C₂H₅NH₃⁺.

pH =

1. c) Use the Henderson–Hasselbalch equation to calculate pH of a buffer solution.

Using the Henderson-Hasselbalch equation, calculate the pH of a buffer solution that is 0.450 M in HSO₃^- and 0.204 M in SO₃^2-.

pH =

1d. ) Calculate buffer pH after adding strong acid or strong base.

Determine the pH change when 0.084 mol HNO₃ is added to 1.00 L of a buffer solution that is 0.489 M in HClO and 0.353 M in ClO^-.

pH after addition − pH before addition = pH change =

Answer 1

1.Calculate the pH of 650. mL of a 0.211-M solution of acetic acid before and after the addition of 0.123 mol of potassium acetate.

pH before addition =
pH after addition =

Ans :-

pH before addition :-

ICE table of the reaction is :

............................CH3COOH (aq)-------------------------> CH3COO^- (aq) ...................+...............H⁺ (aq)

Initial .....................0.211 M...............................................0.0 M...................................................0.0 M

Change .....................-y......................................................+y........................................................+y

Equilibrium ..............(0.211-y) M...........................................y M.....................................................y M

Expression of K_a is :-

K_a = [CH3COO^-].[H⁺] /[CH3COOH]

1.8 x 10^-5 = y²(0.211-y)

y² + 1.8 x 10^-5 y - 3.798 x 10^-6 = 0

Therefore ,

y = 0.00194 M

Therefore,

[H⁺] = 0.00194 m

pH = - log [H⁺]

pH = - log 0.00194 M

pH = 2.71

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pH after addition :

Moles of acetic acid = Molarity x Volume of solution

= 0.211 M x 0.650 L

= 0.13715 mol

Given moles of CH₃COOK = 0.123 mol

By using Henderson- Hasselbalch equation :

pH = pKa + log [CH3COOH]/[CH3COOK]

pH = 4.74 + log 0.123 mol / 0.13715 mol

pH = 4.74 - 0.0473

pH = 4.70