Question

When $CO _{2}$ dissolves in water, the following equilibrium is established.
$CO _{2}+2 H _{2} O \rightleftharpoons H _{3} O ^{+}+ CO _{3}^{-}$
The equilibrium constant is $3.8 \times 10^{-7}$ and at $pH =6.0$, the ratio $\frac{\left[ HCO _{3}{ }^{-}\right]}{\left[ CO _{2}\right]}$ will be

• A. $3.8 \times 10^{-13}$
• B. $3.8 \times 10^{-1}$
• C. $6.0$
• D. $13.4$

Answer 1

Answer: (B)

$CO _{2}+2 H _{2} O \rightleftharpoons H _{3} O ^{+}+ HCO _{3^{-}}$
$K _{ a }=\frac{\left[ H _{3} O \right]+\left[ HCO _{3}^{-}\right]}{\left[ CO _{2}\right]}$
$\frac{\left[ HCO _{3}^{-}\right]}{\left[ CO _{2}\right]}=\frac{ K _{ a }}{\left[ H _{3} O ^{+}\right]}$
$=\frac{3.8 \times 10^{-7}}{10^{-6}}=3.8 \times 10^{-1}$