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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 of the answer using first ionization constant now what is the ph for the Solution containing ha ion
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Here's how to calculate the pH of a 0.150M malonic acid solution using the first ionization constant:
1. Write the equilibrium reaction and the expression for K(1):
Malonic acid (H₂A) is a diprotic acid. Its first ionization is:
H₂A(aq) ⇌ H⁺(aq) + HA⁻(aq)
The equilibrium constant expression is:
K(1) = [H⁺][HA⁻] / [H₂A]
2. Set up an ICE table:
| | H₂A | H⁺ | HA⁻ |
|-------------|---------|--------|---------|
| Initial | 0.150 M | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 0.150-x | x | x |
3. Substitute into the K(1) expression and solve for x:
1.40 x 10⁻³ = (x)(x) / (0.150 - x)
Since K(1) is relatively small, we can assume x
1.40 x 10⁻³ = x² / 0.150
x² = 2.1 x 10⁻⁴
x = [H⁺] ≈ 0.0145 M
4. Calculate the pH:
pH = -log[H⁺] = -log(0.0145) ≈ 1.84
Therefore, the pH of the 0.150M malonic acid solution, considering only the first ionization, is approximately 1.84.
To calculate the pH for the solution containing HA⁻, you would need to consider the second ionization and use the second ionization constant, K(2).
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