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Calculate the pH of 0.10 M of (CH3)2NH4

1 Answer

1 vote

Answer:

5.86

Explanation:

The equation for the equilibrium is

(CH₃)₂NH₂⁺ + H₂O ⇌ H₃O⁺ + (CH₃)₂NH; pKₐ= 10.73

For ease of typing, let’s rewrite the equation as

HA + H₂O ⇌ H₃O⁺ + A⁻

===============

Set up an ICE table:

HA + H₂O ⇌ H₃O⁺ + A⁻

I: 0.10 0 0

C: -x +x +x

E: 0.10-x x x

===============

Calculate Kₐ

Kₐ = 10^(-pKₐ) =10^(-10.73) = 1.86 × 10⁻¹¹

===============

Solve the Kₐ expression

Kₐ = {[H₃O⁺][A⁻]}/[HA] = 1.86 × 10⁻¹¹ Substitute values

(x × x)/(0.10-x) = 1.86 × 10⁻¹¹ Combine like terms

x²/(0.10-x) Check that x ≪ 0.10

0.10/(1.86 × 10⁻¹¹) = 5.37 × 10⁹ > 400 x ≪0.10; Ignore x

x²/0.10 = 1.86 × 10⁻¹¹ Cross-multiply

x² = 0.10 × 1.86 × 10⁻¹¹ Complete the multiplication

x² = 1.86 × 10⁻¹² Take the square root of both sides

x = [H₃O⁺] = 1.36 × 10⁻⁶ Take the negative log of each side

-log[H₃O⁺] = pH = -log(1.36 × 10⁻⁶) Complete the operation

pH = 5.86

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