Calculate the ph of 0.10 m of (ch3)2nh4

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