Question

Both $HCOOH$ and $ CH_{3}COOH$ solutions have equal pH. If $ \frac{K_{1}}{K_{2}}$ (ratio of acid ionization constants) of these acids is $4$, their molar concentration ratio will be

• A. 2
• B. 0.5
• C. 4
• D. 0.25

Answer 1

Answer: (D)

As $pH$ of $HCOOH$ sol. $= pH$ of $CH _{3} COOH$ sol.
$\left[ H ^{+}\right]$ in $HCOOH$ sol. $=\left[ H ^{+}\right]$in $CH _{3} COOH$ sol.
$K_{1}=\frac{\left[ HCOO ^{-}\right]\left[ H ^{+}\right]}{[ HCOOH ]}$
$=\frac{\left[ H ^{+}\right]^{2}}{[ HCOOH ]}$
$K_{2}=\frac{\left[ CH _{3} COO ^{-}\right]\left[ H ^{+}\right]}{\left[ CH _{3} COOH \right]}$
$=\frac{\left[ H ^{+}\right]^{2}}{\left[ CH _{3} COOH \right]}$
$\Rightarrow K_{1}[ HCOOH ]=K_{2}\left[ CH _{3} COOH \right]$
$\therefore \frac{[ HCOOH ]}{\left[ CH _{3} COOH \right]}=\frac{K_{2}}{K_{1}}$
$=\frac{1}{4}=0.25$