Question

What % of the carbon in the H2CO3 – HCO3– buffer should be in the form of HCO3– so as to have a neutral solution?

(Ka = 4 × 10–7)

• A. 20 %
• B. 40 %
• C. 60 %
• D. 80%

Answer 1

Answer: (D)

80%

Answer 2

pH = pKa + log $\frac{\lbrack HCO_{3}^{-}\rbrack}{\lbrack H_{2}CO_{3}\rbrack}$ $\Rightarrow$ 7 = 7 – log 4 + log $\frac{\lbrack HCO_{3}^{-}\rbrack}{\lbrack H_{2}CO_{3}\rbrack}$ $\Rightarrow$$\frac{\lbrack HCO_{3}^{-}\rbrack}{\lbrack H_{2}CO_{3}\rbrack}$= 4

% of Carbon in the form of HCO3–

=$\frac{\lbrack HCO_{3}^{-}\rbrack}{\lbrack HCO_{3}^{-}\rbrack + \lbrack H_{2}CO_{3}\rbrack}$ × 100 = $\frac{4}{1 + 4}$× 100 = 80 %