What is the pH of a 0.15 M NH4OH solution with a Kb of 1.78 x 10^-5? a. 11.21
b. 11.93
c. 12.09
d. 10.85

pH value is the measure of acidic or base properties of a solution.

The answer to the given question is a. 11.21. It is explained as follows.

The range of pH value is between 0 to 14 in which value 7 is the neutral pH which is associate with water.

The water has no dominant properties of acid neither base.

The pH value below 7 is for acidic properties and for above 7 is associate with base properties.

The pH value of 0.15 M NH4OH solution is 11.21 as it displays strong base properties.

The correct answer is a. 11.21.

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The correct option is;

a. 11.21

Explanation:

The chemical equation for the reaction is;

NH₄OH ↔ NH₄⁺ + OH⁻

At the start of the reaction, we have;

   NH₄OH ↔ NH₄⁺ + OH⁻

I    0.15             0        0

C  0.15 - x          +x       +x

The = 1.78 × 10⁻⁵

Therefore, we have;

1.78 × 10⁻⁵ = [NH₄⁺][OH⁻]/NH₄OH = x²/(0.15 - x)

Which gives;

(0.15 - x) 1.78 × 10⁻⁵ = x²

x² + 1.78 × 10⁻⁵x - 2.67× 10⁻⁶ = 0

Factorizing the above equation online, gives;

(x + 0.00164294)(x - 0.00162514) = 0

Therefore;

x = -0.00164294, or x = 0.00162514

We use the positive result, which is x = 0.00162514 = 1.62514 × 10⁻³ = [OH⁻]

Therefore;

pOH = -log[OH⁻] = -log(0.00162514) = 2.789

pH + pOH = 14

Therefore;

pH = 14 - pOH = 14 - 2.789 ≈ 11.21089

The pH of a 0.15 M NH₄OH solution with   = 1.78 × 10⁻⁵ = 11.21.

The pH of a 0.15 M NH₄OH solution with   = 1.78 × 10⁻⁵ is approximately  11.21

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