Question

What is the pH of a $1.0M$ solution of acetic acid? To what volume must $1L$ of this solution be diluted so that the pH of the resulting solution will be twice the original value?

[Given : $K_a=1.8\times {10}^{-5}$]

• A. $2.38,244$
• B. $2.38,122$
• C. $4.76,244$
• D. $4.76,122$

Answer 1

Answer: (A)

The concentration of acid is $1M$.
The expression for the equilibrium constant is $K=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

Substituting values in the above expression, we get-
$1.8\times {10}^{-5}=\frac{x^2}{1-x}$

Since the value of the equilibrium constant is very small, $1-x\simeq 1$
Hence, $1.8\times {10}^{-5}=\frac{x^2}{1}$

$x^2=1.8\times {10}^{-5}$

$x=4.2\times {10}^{-3}$

Therefore, $pH=-log\left[H^{+}\right]=-log4.24\times {10}^{-3}=2.38$

The pH is doubled to $2\times 2.38=4.76$

The new concentration is $\left[H^{+}\right]={10}^{-pH}={10}^{-4.76}=1.734\times {10}^{-5}$

The new volume of the solution is $\frac{4.24\times {10}^{-3}M\times 1L}{1.734\times {10}^{-5}}=244L$

Hence, $1L$ of this solution must be diluted to a volume of $244L$ so that the pH of the resulting solution will be twice the original value.