Question

1 M solution of $\text{ C}{{\text{H}}_{\text{3}}}\text{COOH }$ should be diluted to _________times so that $\text{ pH }$is doubled. (Given $\text{ }{{\text{K}}_{\text{a}}}\text{ (C}{{\text{H}}_{\text{3}}}\text{COOH) = 1}\text{.8}\times \text{1}{{\text{0}}^{-5}}\text{ }$ )
A) 4 times
B) $\text{5}\text{.55 }\times \text{1}{{\text{0}}^{\text{4}}}\text{ }$ Times
C) $\text{5}\text{.55 }\times \text{1}{{\text{0}}^{6}}\text{ }$Times
D) $\text{ 1}{{\text{0}}^{-2}}\text{ }$Times

Answer 1

Hint: The dissociation constant for the weak acid (like acetic acid) is related to the degree of dissociation and concentration of the solution. However, the relation can be further modified to calculate the hydrogen ion concentration of the aqueous solution of the weak acid whose dissociation constant is known. The relation is as shown below,
$\text{ }\left[ {{\text{H}}^{\text{+}}} \right]\text{ = }\sqrt{c{{\text{K}}_{\text{a}}}}\text{ }$
Where ${{\text{K}}_{\text{a}}}$ dissociation constant, c is the concentration, and $\left[ {{\text{H}}^{\text{+}}} \right]$ is hydrogen ion concentration.

Complete step by step solution:
Acetic acid is a weak acid. Its dissociation constant $\text{ }{{\text{K}}_{\text{a}}}\text{ }$ is determined by the equation which related the dissociation constant $\text{ }{{\text{K}}_{\text{a}}}\text{ }$with the concentration of acetic acid and the degree of dissociation. the equation is as follows,
$\text{ }{{\text{K}}_{\text{a}}}\text{ = c}{{\alpha }^{\text{2}}}\text{ OR }\alpha \text{ = }\sqrt{\dfrac{{{\text{K}}_{\text{a}}}\text{ }}{\text{c}}}\text{ }$ (1)
The above equation can be rewritten in terms of the hydrogen ion concentration. Such that hydrogen ion concentration in the solution is equal to the $\text{ c}\alpha \text{ }$ . Thus the relation is stated as follows,
$\text{ }\left[ {{\text{H}}^{\text{+}}} \right]\text{ = c}\alpha \text{ = c}\sqrt{\dfrac{{{\text{K}}_{\text{a}}}\text{ }}{\text{c}}}\text{ = }\sqrt{c{{\text{K}}_{\text{a}}}}\text{ }$ (2)
Take a negative logarithm of the equation (2) .we have,
$\text{ }-\log \left[ {{\text{H}}^{\text{+}}} \right]\text{ = }-\log \text{ }\sqrt{c{{\text{K}}_{\text{a}}}}\text{ = }\dfrac{1}{2}\left[ -\log \text{ c +}(-\text{ log }{{\text{K}}_{\text{a}}})\text{ } \right]\text{ }$
Since we know that, $\text{ }-\log \left[ {{\text{H}}^{\text{+}}} \right]\text{ = pH }$ and$\text{ }-\text{ log }{{\text{K}}_{\text{a}}}\text{ = p}{{\text{K}}_{\text{a}}}\text{ }$, we have
$\text{ pH = }\dfrac{1}{2}\left[ \text{p}{{\text{K}}_{\text{a}}}-\log \text{ c } \right]\text{ }$ (3)
Let's substitute the values in equation (3). We get,
$\text{ pH = }\dfrac{1}{2}\left[ \text{p}{{\text{K}}_{\text{a}}}-\log \text{ 1 } \right]\text{ = }\dfrac{\text{p}{{\text{K}}_{\text{a}}}}{2}\text{ }$ (4)
We are interested in the value of the concentration when the value of $\text{ pH }$ doubled. We already know the value of $\text{ pH }$in terms of $\text{ p}{{\text{K}}_{\text{a}}}\text{ }$ . Thus to get the twice of $\text{ pH }$multiply the equation (4) by 2.the value of new $\text{pH }\!\!'\!\!\text{ }$ is,
$\text{ pH }\!\!'\!\!\text{ =2}\times \text{pH = 2}\times \dfrac{\text{p}{{\text{K}}_{\text{a}}}}{2}=\text{ p}{{\text{K}}_{\text{a}}}$
Now substitute these values in equation (3), so that we can determine the concentration $\text{ C }\!\!'\!\!\text{ }$ which is a concentration of acetic acid required for the double value of $\text{ pH }$ the solution. We have,
\[\text{ pH }\!\!'\!\!\text{ = p}{{\text{K}}_{\text{a}}}\text{= }\dfrac{1}{2}\left[ \text{p}{{\text{K}}_{\text{a}}}-\log \text{ C }\!\!'\!\!\text{ } \right]\text{ }\]
On further simplifying we have,
$\begin{align}
& \text{ 2p}{{\text{K}}_{\text{a}}}-\text{p}{{\text{K}}_{\text{a}}}\text{= }\left[ -\log \text{ C }\!\!'\!\!\text{ } \right]\text{ } \\
& \Rightarrow \text{p}{{\text{K}}_{\text{a}}}\text{ = }-\log \text{ C }\!\!'\!\!\text{ } \\
& but,\text{ p}{{\text{K}}_{\text{a}}}\text{ = }-\log {{\text{K}}_{\text{a}}} \\
& \Rightarrow \text{ }-\log {{\text{K}}_{\text{a}}}\text{ = }-\log \text{ C }\!\!'\!\!\text{ } \\
& \therefore \text{ }{{\text{K}}_{\text{a}}}\text{ = C }\!\!'\!\!\text{ } \\
\end{align}$
We know that dissociation constant for the acetic acid is $\text{ }{{\text{K}}_{\text{a}}}\text{ = 1}\text{.8 }\times \text{1}{{\text{0}}^{-5}}\text{ }$ , therefore dilution ‘x’ times would be equal to,
$\text{ dilution }\!\!'\!\!\text{ x }\!\!'\!\!\text{ = }\dfrac{1}{C'}\text{ = }\dfrac{1}{1.8\times {{10}^{-5}}}=\text{ 5}\text{.55 }\times \text{1}{{\text{0}}^{\text{4}}}\text{ times }$
Thus to double the $\text{ pH }$ 1 M solution of acetic acid we need to dilute it by the $\text{5}\text{.55 }\times \text{1}{{\text{0}}^{\text{4}}}$ times.

Hence, (B) is the correct option.

Note: The negative logarithm taking a step is essential in the determination of $\text{ pH }$as direct formula cannot be employed to determine the $\text{ pH }$. This will direct us to the hydrogen concentration. The hydroxyl ion concentration of the weak base can also be written in terms of dissociation constant f as follows,$\text{ }\left[ \text{O}{{\text{H}}^{-}} \right]\text{ = }\sqrt{c{{\text{K}}_{\text{b}}}}\text{ }$. where ${{\text{K}}_{\text{b}}}$ is the dissociation constant of a weak base such as ammonium hydroxide.