Q.The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10⁻³M. what is its pH ?

Concept & Intuition

pH is a measure of hydrogen ion concentration, but in a strong base like NaOH, the hydroxide ion concentration directly determines the pH via the water dissociation constant. Since NaOH dissociates completely, the mass needed is simply the molar mass times the required moles of OH⁻. The key is to convert pH to pOH first, because pOH gives [OH⁻] directly.

Step-by-step solution

1. Relate pH and pOH

   At 25 °C, pH + pOH = 14.

   Given pH = 12, so pOH = 14 − 12 = 2.

2. Find [OH⁻] from pOH

   pOH = −log₁₀[OH⁻], so [OH⁻] = 10⁻² M = 0.01 mol/L.

3. Determine moles of NaOH needed

   NaOH is a strong base: NaOH → Na⁺ + OH⁻ completely.

   Thus, [NaOH] = [OH⁻] = 0.01 M.

   For 1.0 L solution: moles of NaOH = 0.01 mol/L × 1.0 L = 0.01 mol.

4. Convert moles to mass

   Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.

   Mass = 0.01 mol × 40 g/mol = 0.40 g.

The key concept here is the Henderson–Hasselbalch equation for a buffer solution. A buffer containing a weak acid (acetic acid, CH₃COOH) and its conjugate base (acetate, CH₃COO⁻) resists pH change when small amounts of strong base are added. The equation is:

When NaOH is added, it reacts completely with the acid:

So the moles of acid decrease and the moles of acetate increase by the same amount. The trick is to track these changes precisely.

1. Find moles of NaOH added

   Molar mass of NaOH = 40 g/mol.

   Moles of NaOH = $\frac{8}{40} = 0.2$ mol.

2. Initial moles in the buffer

   We have 1 mole of acetic acid (HA) and 1 mole of sodium acetate (A⁻) in 1.0 L solution.

   So initially: $n_{\text{HA}} = 1.0$, $n_{\text{A}^-} = 1.0$.

3. Reaction with NaOH

   The 0.2 mol OH⁻ will consume 0.2 mol of HA and produce 0.2 mol of A⁻.

   After reaction:

4. Concentrations in the same volume Volume remains 1.0 L, so concentrations are numerically equal to moles:

5. Apply Henderson–Hasselbalch

Now $\log 1.5 = \log \frac{3}{2} = \log 3 - \log 2 = 0.47 - 0.30 = 0.17$

So:

We have a mixture of a weak acid (acetic acid) and its conjugate base (acetate from sodium acetate). That’s a classic buffer. The pH of a buffer is given by the Henderson–Hasselbalch equation:

The key here is that we are mixing two solutions and then diluting to a final volume. The ratio of concentrations in the final solution is the same as the ratio of moles (since both are in the same final volume). So we don’t need the final concentrations — just the moles of acid and base.

Let’s work through it step by step.

1. Find moles of acetic acid (HA)

   Volume = 10 mL = 0.010 L, concentration = 1.0 M

   Moles of HA = $0.010 \times 1.0 = 0.010$ mol.

2. Find moles of acetate (A⁻)

   Volume = 20 mL = 0.020 L, concentration = 0.5 M

   Moles of A⁻ = $0.020 \times 0.5 = 0.010$ mol.

3. Dilution effect

   The mixture is diluted to 100 mL. Both species are in the same final volume, so the ratio of their concentrations is exactly the ratio of their moles:

4. Apply Henderson–Hasselbalch

The key concept here is the buffer solution. A mixture of a weak acid (acetic acid, $\mathrm{CH_3COOH}$) and its salt with a strong base (sodium acetate, $\mathrm{CH_3COONa}$) resists drastic pH changes. The salt fully dissociates, providing a reservoir of the conjugate base ($\mathrm{CH_3COO^-}$). The weak acid remains mostly undissociated. The pH of such a mixture is governed by the equilibrium:

Because the acid is weak and the conjugate base is present in significant amount from the salt, we can use the Henderson–Hasselbalch equation, which is derived directly from the acid dissociation constant expression:

This works because the equilibrium concentration of the acid is approximately its initial concentration (very little dissociates), and the equilibrium concentration of the conjugate base is approximately the initial concentration from the salt (the small amount from acid dissociation is negligible).

Let’s apply it step by step.

1. Identify the given values.

   • $[\mathrm{CH_3COOH}] = 0.2\ \mathrm{M}$ (weak acid)
   • $[\mathrm{CH_3COONa}] = 0.1\ \mathrm{M}$ (salt, gives $[\mathrm{CH_3COO^-}] = 0.1\ \mathrm{M}$)
   • $K_a = 1.8 \times 10^{-5}$, so $\mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74$

2. Plug into the Henderson–Hasselbalch equation.

3. Simplify the ratio.

So:

4. Evaluate the logarithm.

   $\log(0.5) = \log(1/2) = -\log 2 \approx -0.3010$

5. Compute the pH.

The closest option to 4.44 is 4.43 (option B).

Given: a $0.1\ \mathrm{M}$ monobasic acid with $\mathrm{pH}=2$.

Degree of dissociation.

Effective (osmotically active) concentration.

Osmotic pressure.