Concept Explanation: Calculating Ka from pH and Concentration

Given: pH = 3 for a 0.1 M solution of acid HQ. We need to find the ionization constant, Ka.

For a weak acid, the dissociation is: $\mathrm{HQ}\rightleftharpoons H^{+}+Q^{-}$

The pH is defined as: $\mathrm{pH}=-\log \left[H^{+}\right]$

Step-by-Step Solution:

Step 1: Find [H⁺] from pH

Given pH = 3, so $-\log \left[H^{+}\right]=3$

Therefore, $\left[H^{+}\right]={10}^{-3}M=0.001M$

Step 2: Set up the Ka expression

For the reaction $\mathrm{HQ}\rightleftharpoons H^{+}+Q^{-}$, the Ka is:

$K_a=\frac{\left[H^{+}\right]\left[Q^{-}\right]}{\left[\mathrm{HQ}\right]}$

Step 3: Determine equilibrium concentrations

Initial [HQ] = 0.1 M

Let x = [H⁺] at equilibrium = 0.001 M (from Step 1)

Since [H⁺] = [Q⁻] = x, and the change in [HQ] is -x

So, at equilibrium:

$\left[H^{+}\right]=x=0.001M$

$\left[Q^{-}\right]=x=0.001M$

$\left[\mathrm{HQ}\right]=0.1-x\approx 0.1M$ (since x << 0.1, approximation is valid)

Step 4: Substitute into Ka expression

$K_a=\frac{\left(x\right)\left(x\right)}{0.1}=\frac{x^2}{0.1}=\frac{(0.001{)}^2}{0.1}=\frac{0.000001}{0.1}=0.00001=1\times {10}^{-5}$

Final Answer: The value of the ionization constant Ka is $1\times {10}^{-5}$

Related Topics:

• Weak Acid Dissociation
• pH and pKa Calculations
• Equilibrium Constants
• Acid-Base Chemistry

Important Formulae:

$\mathrm{pH}=-\log \left[H^{+}\right]$

$\left[H^{+}\right]={10}^{-\mathrm{pH}}$

For weak acid HA: $K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[\mathrm{HA}\right]}$

For dilute solutions: $\left[H^{+}\right]\approx \sqrt{K_a\times C}$ where C is initial concentration