Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0136 M solution. The pH of the resulting solution is 2.45. Calculate the Ka for the acid.

Answer 1

Answer: $K_a=1.2\times 10^{-6}$

Explanation: $HA\rightarrow H^++A^-$

initially conc.         c                              0         0

At eqm.    $c(1-\alpha)$      $c\alpha$        $c\alpha$

The expression for dissociation constant is,

$K_a=\frac{c\alpha\times c\alpha}{c(1-\alpha)}$

And,

$[H^+]=c\alpha$

$pH=-log[H^+]=2.45$

$H^+=3.5\times 10^{-3}$

$3.5\times 10^{-3}=0.0136\alpha$

$\alpha=0.26$

Now put all the given values in this expression, we get

$K_a=\frac{(0.0136\times 0.26)^2}{0.0136(1-0.26)}$

$K_a=1.2\times 10^{-6}$

Answer 2

An acid that is capable of donating a single proton or the hydrogen ion is called monoprotic acid. The Ka for the acid will be $1.2 \times 10^{-6}$.

What is the acid dissociation constant?

Ka or the acid dissociation constant is the value that differentiates the weak and strong acids from each other and tells about the dissociation of the acid in the solution.

The acid dissociation can be shown as:

$\rm HA \rightarrow H^{+} + A^{-}$

The initial concentration of the reactant is c and of products is 0. While at equilibrium the concentration of the reactant is $\rm c\;(1- \alpha )$ and of products is $\rm c \;\alpha$.

The dissociation constant can be shown by the equation:

$\rm K_{a} = \dfrac{c \;\alpha \times c\; \alpha}{\rm c(1- \alpha)}$

As we know that the $\rm [H^{+}] = \rm c \;\alpha.$

And, the formula for pH is:

$\rm pH = -log [H^{+}]$

Given,

• pH = 2.45
• Concentration of hydrogen ion = $3.5 \times 10^{-3}$

Substituting values in the above equation:

$\begin{aligned} 3.5 \times 10^{-3} &= 0.0136\; \alpha\\\\&= 0.26\end{aligned}$

Now substituting values in the Ka equation:

$\begin{aligned}\rm K_{a} &= \dfrac{(0.0136 \times 0.26)^{2}}{0.0136(1-0.26)}\\\\&= 1.2 \times 10 ^{-6}\end{aligned}$

Therefore, $1.2 \times 10 ^{-6}$ is the Ka or the acid dissociation constant.

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