Ionic EquilibriumHard

Question

The active ingredient in aspirin is acetyl salicylic acid

with Ka = 4.0 × 10–9. The pH of the solution obtained by dissolving two aspirin tablets (containing 0.36 g of acetyl salicylic acid in each tablet) in 250 ml of water is (log 2 = 0.3)

Options

A.5.1
B.8.9
C.10.2
D.5.25

Solution

Conc. of aspirin = $\frac{2 \times 0.36/180}{0.250} = 0.016\text{ M}$

Now, $\left\lbrack H^{+} \right\rbrack = \sqrt{K_{a}.C} = \sqrt{4 \times 10^{- 9} \times 0.016} = 8 \times 10^{- 6}\text{ M}$

∴ PH = 5.1

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