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Calculate ph of a 0.150m solution of malonic acid. Ionization constant for the acid are K(1)= 1.40 x 10 ^ -3 ; K(2) = 2.2 x 10^-6 give the figure if the answer using first ionization constant
Accepted Answer
The pH of a 0.150M solution of malonic acid, considering only the first ionization constant, can be calculated as follows:
1. Set up an ICE table:
| | H₂C₃H₂O₄ | H⁺ | HC₃H₂O₄⁻ |
|---|---|---|---|
| I | 0.150M | 0 | 0 |
| C | -x | +x | +x |
| E | 0.150-x | x | x |
2. Write the expression for K₁:
K₁ = [H⁺][HC₃H₂O₄⁻] / [H₂C₃H₂O₄] = 1.40 x 10⁻³
3. Substitute the equilibrium concentrations from the ICE table:
1.40 x 10⁻³ = (x)(x) / (0.150 - x)
4. Since K₁ is relatively small, we can assume x
1.40 x 10⁻³ ≈ (x)² / 0.150
5. Solve for x (which represents [H⁺]):
x = √(1.40 x 10⁻³
0.150) ≈ 0.0145 M
6. Calculate the pH:
pH = -log[H⁺] = -log(0.0145) ≈ 1.84
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