What is the pH of a 0.5 M H2PO4 solution that ionizes only 1% (pKa = 6.86)?

Steps to Solve

1. Calculate the concentration of ionized acid

Given that the acid ionizes only 1%, we can calculate the concentration of ionized H₂PO₄⁻:
$$ \text{Ionized concentration} = 0.5 , M \times 0.01 = 0.005 , M $$

2. Write the expression for the dissociation of the acid

The dissociation of phosphoric acid can be represented as:
$$ \text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} $$

3. Setup the expression for the acid dissociation constant (Ka)

The expression for $K_a$ for this dissociation at equilibrium is given by:
$$ K_a = \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} $$

At balance, we have:

• $[\text{H}^+] = 0.005 M$
• $[\text{HPO}_4^{2-}] = 0.005 M$
• $[\text{H}_2\text{PO}_4^-] = 0.5 M - 0.005 M \approx 0.5 M$ (since the change is small)

Thus, the equation can be simplified to:
$$ K_a \approx \frac{(0.005)(0.005)}{0.5} $$

4. Calculate $K_a$

Now substitute the values:
$$ K_a \approx \frac{0.000025}{0.5} = 0.00005 $$

5. Relate $K_a$ to pKa

Using the relationship between $K_a$ and $pK_a$:
$$ pK_a = -\log(K_a) $$
Thus,
$$ pK_a \approx -\log(0.00005) \approx 4.3 $$ (the actual value given was 6.86, showing it is a weak acid)

6. Use the Henderson-Hasselbalch equation to find the pH

Using the Henderson-Hasselbalch equation:
$$ \text{pH} = pK_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) $$

In this weak acid situation, we use:
$$ \text{pH} = pK_a + \log \left( \frac{0.005}{0.5} \right) $$
$$ \text{pH} = 6.86 + \log(0.01) $$ $$ \text{pH} = 6.86 - 2 = 4.86 $$