The answer is B (\(1.6 \times 10^{-4}\)).
\(K_\text{a} = \frac{[\text{HCO}_3^-][\text{H}^+]}{[\text{H}_2\text{CO}_3]}\), so \(\frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]} = \frac{[\text{H}^+]}{K_\text{a}}\).
At pH 7.4, \([\text{H}^+] = 10^{-7.4} = 3.98 \times 10^{-8}\) mol dm\(^{-3}\).
\(\frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]} = \frac{3.98 \times 10^{-8}}{2.5 \times 10^{-4}} = 1.6 \times 10^{-4}\)
Check with \(\text{pH} = \text{p}K_\text{a} - \lg\frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]}\), where \(\text{p}K_\text{a} = -\lg(2.5 \times 10^{-4}) = 3.60\): pH = 3.60 + 3.80 = 7.40.
• A is wrong: a ratio of \(3.4 \times 10^{-5}\) gives pH = 3.60 + 4.47 = 8.07, not 7.4.
• C is wrong: \(6.3 \times 10^3\) is the inverse ratio, \(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\); used as asked it gives pH = 3.60 - 3.80 = -0.20.
• D is wrong: a ratio of \(3.0 \times 10^4\) gives pH = 3.60 - 4.48 = -0.88, not 7.4.
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