Calculate the pH of a 0.2M solution of HC2H3O2 (Ka = 1.8 x 10-5).2.720.695.444.74
Question
Calculate the pH of a 0.2M solution of HC2H3O2 (Ka = 1.8 x 10^-5).
Options:
- 2.72
- 0.69
- 5.44
- 4.74
Solution
To calculate the pH of a 0.2M solution of HC2H3O2, we can use the formula for the ionization constant of water (Ka) and the definition of pH. Here are the steps:
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Write the ionization equation for acetic acid (HC2H3O2): HC2H3O2 ⇌ H+ + C2H3O2-
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Write the expression for the ionization constant (Ka): Ka = [H+][C2H3O2-] / [HC2H3O2]
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Since the solution is 0.2M, initially, before any ionization, we have 0.2M of HC2H3O2, and 0M of H+ and C2H3O2-. As ionization occurs, let's say x M of HC2H3O2 ionizes. So, we have x M of H+ and C2H3O2-, and 0.2-x M of HC2H3O2.
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Substitute these values into the Ka expression: 1.8 x 10^-5 = x^2 / (0.2 - x)
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Since Ka is very small, we can assume that x is much smaller than 0.2, so the equation simplifies to: 1.8 x 10^-5 = x^2 / 0.2
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Solve for x, which represents the [H+]: x = sqrt(1.8 x 10^-5 * 0.2) = 1.8973665961010275e-3 M
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Calculate the pH using the definition of pH = -log[H+]: pH = -log(1.8973665961010275e-3) = 2.72
So, the pH of the 0.2M solution of HC2H3O2 is approximately 2.72.
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