Question

Given equilibrium constants: H$_2$CO$_3$ ⇌ 2H$^+$ + CO$_3^{2–}$, $K = 2 \times 10^{-17}$
H$_2$CO$_3$ ⇌ H$^+$ + HCO$_3^-$, $K = 4 \times 10^{-7}$
Then the equilibrium constant for: HCO$_3^-$ ⇌ H$^+$ + CO$_3^{2–}$ is

• A. $(4 \times 10^{-7}) + (2 \times 10^{-17})$
• B. $4 \times 10^{-7} \cdot 2 \times 10^{-17}$
• C. $5 \times 10^{-11}$
• D. $8 \times 10^{-24}$

Hint:

When combining equilibria, divide constants when reversing steps: $K = \dfrac{K_1}{K_2}$.

Answer 1

The Correct Option is C

Step 1: $K_1$ for H$_2$CO$_3$ ⇌ 2H$^+$ + CO$_3^{2–}$ = $2 \times 10^{-17}$
Step 2: $K_2$ for H$_2$CO$_3$ ⇌ H$^+$ + HCO$_3^-$ = $4 \times 10^{-7}$
To get equilibrium constant for: HCO$_3^-$ ⇌ H$^+$ + CO$_3^{2–}$ = $\dfrac{K_1}{K_2}$
$= \dfrac{2 \times 10^{-17}}{4 \times 10^{-7}} = 0.5 \times 10^{-10} = 5 \times 10^{-11}$