When CO 2 dissolves in water, the following equilibrium is established. CO 2+2 H 2 O leftharpoons H 3 O ++ CO 3- The equilibrium constant is 3.8 × 10-7 and at pH =6.0, the ratio ([ HCO 3 -]/[ CO 2]) will be

Question

When CO 2 dissolves in water, the following equilibrium is established. CO 2+2 H 2 O leftharpoons H 3 O ++ CO 3- The equilibrium constant is 3.8 × 10-7 and at pH =6.0, the ratio ([ HCO 3 -]/[ CO 2]) will be (A) 3.8 × 10-13 (B) 3.8 × 10-1 (C) 6.0 (D) 13.4

Tardigrade Admin · Accepted Answer

CO 2+2 H 2 O leftharpoons H 3 O ++ HCO 3- K a =([ H 3 O ]+[ HCO 3-]/[ CO 2]) ([ HCO 3-]/[ CO 2])=( K a /[ H 3 O +]) =(3.8 × 10-7/10-6)=3.8 × 10-1

Q. When $C O_{2}$ dissolves in water, the following equilibrium is established.
$C O_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + C O_{3}^{-}$
The equilibrium constant is $3.8 \times 1 0^{- 7}$ and at $p H = 6.0$ , the ratio $\frac{[ H C O _{3} ^{-} ]}{[ C O _{2} ]}$ will be

$3.8 \times 1 0^{- 13}$

$3.8 \times 1 0^{- 1}$

$6.0$

$13.4$

$C O_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + H C O_{3^{-}}$
$K_{a} = \frac{[ H _{3} O ] + [ H C O _{3}^{-} ]}{[ C O _{2} ]}$
$\frac{[ H C O _{3}^{-} ]}{[ C O _{2} ]} = \frac{K _{a}}{[ H _{3} O ^{+} ]}$
$= \frac{3.8 \times 1 0 ^{- 7}}{1 0 ^{- 6}} = 3.8 \times 1 0^{- 1}$

Q. When CO2​ dissolves in water, the following equilibrium is established. CO2​+2H2​O⇌H3​O++CO3−​ The equilibrium constant is 3.8×10−7 and at pH=6.0, the ratio [CO2​][HCO3​−]​ will be

3.8×10−13

3.8×10−1

6.0

13.4