Question

In a $20ml$ $0.4{\text{M}} - {\text{HA}}$ solution, $80ml$ water is added. Assuming volume to be additive, the $pH$ off final solution is $\left( {{{\text{K}}_{\text{a}}}{\text{ of HA}} = 4 \times {{10}^{ - 7}},\log 2 = 0.3} \right)$
A) $5.30$
B) $4.30$
C)$3.50$
D) $3.70$

Answer 1

Hint:We must remember that the concentration of hydrogen ion in the solution is termed as $pH$ of the solution. It is the general way to determine the strength of base/acid. The $pH$ value ranges from $0 - 14$. For acid the value of $pH$ is $ < 7$, the $pH$ value base is $ > 7$, for the neutral molecule the value of $pH$ is equal to $0$.

Complete step by step answer:
The molarity of the acid and base titration is calculated using the relation.
${M_1}{V_1} = {M_1}{V_2}$
Where,
The molarity of the acidic solution is ${M_1}$.
The volume of the acidic solution is ${V_1}$.
The molarity of the basic solution is ${M_2}$.
The volume of the basic is ${V_2}$.
Given,
The volume of water is $80ml$
The volume of the solution is $20ml$.
The molarity of the solution is $0.4{\text{M}}$.
The total volume of the solution is $100ml$.
The constant ${K_a}$ of the solution is $4 \times {10^{ - 2}}$$4 \times {10^{ - 2}}$.
First, calculate the molarity of the solution.
${M_2} = \dfrac{{{M_1}{V_1}}}{{{V_2}}}$
Substituting the values we get,
$ \Rightarrow $${M_2} = \dfrac{{0.4 \times 20}}{{100}}$
The molarity of solution is $0.08{\text{M}}$.
Write the dissociation equation of the reaction.
$HA + {H_2}O\xrightarrow{{}}{H_3}{O^ + } + {A^ - }$
The dissociation constant of the reaction ${K_a}$ is written as,
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$
Let us imagine the concentration of \[\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]\] as x.
$4 \times {10^{ - 7}} = \dfrac{{{x^2}}}{{0.08 - x}}$
$ \Rightarrow $${x^2} = 4 \times {10^{ - 7}} \times 0.08$
$ \Rightarrow $$x = 1.78 \times {10^{ - 4}}$
The concentration of Hydrogen is $1.78 \times {10^{ - 4}}$.
We can calculate the $pH$ of the solution is,
$pH = - \log \left[ {{H^ + }} \right] = 3.75$
The $pH$ of the solution is $3.75$ which is closely related to option D.
Therefore, the option D is correct.

Note: We must remember that the \[pH\] value is decided from the negative logarithm of this concentration and is employed to point the acidic, basic, or neutral character of the substance you're testing. The \[pH\] indicators are weak acids which exist as natural dyes and indicate the concentration of hydrogen ions during a solution via color change.