pH of Weak Acids¶
Example:¶
pH of 0.30 M HF solution (\(K_a=7.1\times10^{−4)}\))
| HF | \(\ce{<=>}\) | \(\ce{H+}\) | + | \(\ce{F-}\) | |
|---|---|---|---|---|---|
| I | 0.30 | 0 | 0 | ||
| C | \(−x\) | \(+x\) | \(+x\) | ||
| E | \(0.3 −x\) | \(x\) | \(x\) |
Quadratic Method:¶
\[
\begin{align}
\ce{K_a&=\frac{[H+][F−]}{[HF]}=\frac{x^2}{0.3-x}=7.1\times 10^{−4}}\\
x^2&=−7.1\times10^{−4}x+0.000213\\
0&=−x^2−7.1\times10^{−4}x+0.000213
\end{align}
\]
solve for \(\pm x\)
Assumption Method:¶
Note
Simplifies the process to:
1. \(x\approxeq\sqrt{K_x\times C_i}\)
2. Check
3. Solve
Assume \(0.3−x\approxeq0.3\)
\[
\begin{align}
\ce{K_a&=\frac{[H+ ][F− ]}{[HF]}=\frac{x^2}{0.3}=7.1\times10^{−4}}\\
x^2&=0.000213\\
\sqrt{0.000213}&=0.146\\
\end{align}
\]
Check: \(\frac{0.0146}{0.3}×100=4.9\%\) \(4.9\%<5\%\) OK!
Solve: \(pH=−log(0.0146)=1.84\)