Common-Ion Effect — Weak Acid/Base with Its Salt (Theory)
A common ion supplied by a salt suppresses the ionization of a weak acid or a weak base.
In an acid mixture the salt provides \( \mathrm{A^-} \) (conjugate base of HA);
in a base mixture the salt provides \( \mathrm{BH^+} \) (conjugate acid of B).
The equilibrium shift reduces the amount that the weak species ionizes, changing
\( [\mathrm{H_3O^+}] \) or \( [\mathrm{OH^-}] \), and hence the pH/pOH.
1) Weak acid HA with its salt NaA
HA + H2O ⇌ A− + H3O+
Suppose the initial concentrations are \( C_{\text{weak}} \) for HA and \( C_{\text{salt}} \) for \( \mathrm{A^-} \).
Using an ICE table with change \( \pm x \):
| HA | A⁻ | H₃O⁺ |
|---|
| Initial | \( C_{\text{weak}} \) | \( C_{\text{salt}} \) | 0 |
| Change | \( -x \) | \( +x \) | \( +x \) |
| Equilibrium | \( C_{\text{weak}}-x \) | \( C_{\text{salt}}+x \) | \( x \) |
\[
\begin{aligned}
K_a
&= \frac{[\mathrm{A^-}][\mathrm{H_3O^+}]}{[\mathrm{HA}]}
= \frac{x\cdot\big(C_{\text{salt}}+x\big)}{C_{\text{weak}}-x}.
\end{aligned}
\]
Rearranging gives a quadratic for \(x=[\mathrm{H_3O^+}]\):
\[
\begin{aligned}
x^{2} + \big(C_{\text{salt}} + K_a\big)\,x - K_a\,C_{\text{weak}} &= 0,\\[4pt]
x &= \frac{-\big(C_{\text{salt}} + K_a\big)
+ \sqrt{\big(C_{\text{salt}} + K_a\big)^{2} + 4\,K_a\,C_{\text{weak}}}}{2}.
\end{aligned}
\]
Once \(x\) is known:
\[
\mathrm{pH} = -\log_{10}\!\big([\mathrm{H_3O^+}]\big) = -\log_{10}(x), \qquad
\mathrm{pOH} = 14.00 - \mathrm{pH}.
\]
2) Weak base B with its salt BH+Cl−
B + H2O ⇌ BH+ + OH−
With initial concentrations \( C_{\text{weak}} \) (B) and \( C_{\text{salt}} \) (\( \mathrm{BH^+} \)),
the ICE table is analogous; letting \( x=[\mathrm{OH^-}] \):
\[
\begin{aligned}
K_b
&= \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}
= \frac{x\cdot\big(C_{\text{salt}}+x\big)}{C_{\text{weak}}-x}.
\end{aligned}
\]
\[
\begin{aligned}
x^{2} + \big(C_{\text{salt}} + K_b\big)\,x - K_b\,C_{\text{weak}} &= 0,\\[4pt]
x &= \frac{-\big(C_{\text{salt}} + K_b\big)
+ \sqrt{\big(C_{\text{salt}} + K_b\big)^{2} + 4\,K_b\,C_{\text{weak}}}}{2}.
\end{aligned}
\]
Then
\[
\mathrm{pOH} = -\log_{10}\!\big([\mathrm{OH^-}]\big) = -\log_{10}(x), \qquad
\mathrm{pH} = 14.00 - \mathrm{pOH}.
\]
3) Practical notes
- \( K_w = 1.0\times 10^{-14} \) at 25 °C; it changes with temperature.
- If \( C_{\text{salt}} \) is much larger than \( K_a \) (acid case) or \( K_b \) (base case),
\(x\) is typically small relative to \( C_{\text{weak}} \) and \( C_{\text{salt}} \). The calculator always solves the exact quadratic, so no approximation is required.
- Setting \( C_{\text{salt}}=0 \) reduces to the standard weak acid/base equilibrium.
- All concentrations are molarity (mol·L\(^{-1}\)). Use consistent units throughout.
4) Quick numeric check (like the textbook example)
For \( C_{\text{weak}}=0.100\ \mathrm{M} \) acetic acid, \( C_{\text{salt}}=0.100\ \mathrm{M} \) sodium acetate, and
\( K_a=1.8\times 10^{-5} \):
\[
\begin{aligned}
x &= \frac{-\big(0.100 + 1.8\times 10^{-5}\big)
+ \sqrt{\big(0.100 + 1.8\times 10^{-5}\big)^{2} + 4\cdot 1.8\times 10^{-5}\cdot 0.100}}{2}
\;\approx\; 1.8\times 10^{-5}\ \mathrm{M},\\
\mathrm{pH} &= -\log_{10}\!\big(1.8\times 10^{-5}\big) \approx 4.74,\quad
[\mathrm{A^-}] \approx 0.100\ \mathrm{M},\quad
[\mathrm{HA}] \approx 0.100\ \mathrm{M}.
\end{aligned}
\]
The calculator implements these exact relations and reports \( [\mathrm{H_3O^+}] \) or \( [\mathrm{OH^-}] \),
the conjugate concentrations, and the resulting pH and pOH.
