First slide

Ionization constants of weak acids, PH of weak acids

Question

When CO₂ is bubbled in excess of water, the following equilibrium is established. ${CO}_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + {HCO}_{3}^{-}$
K_c = 3.8×10^–7, pH = 6. What would be the $\frac{[{HCO}_{3}^{-}]}{[{CO}_{2}]}$ ?

Easy

A

6
B

0.0038
C

0.038
D

0.38

Solution

Given equilibrium is , ${CO}_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + {HCO}_{3}^{-}$
$Kc = \frac{[H_{3} O^{+}] [{HCO}_{3}^{-}]}{[{CO}_{2}]}$
$\Rightarrow 3.8 \times 10^{- 7} = 10^{- 6} \times \frac{[{HCO}_{3}^{-}]}{[{CO}_{2}]}$
$\Rightarrow \frac{[{HCO}_{3}^{-}]}{[{CO}_{2}]} = 3.8 \times 10^{- 1}$

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