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A $0.1M$ solution of weak acid $HA$ is $1%$ dissociated at $298K$. The new $pH$ when $0.2M$ of $NaA$ is added to it is $530\times {{10}^{-x}}$, then what will be the value of $x$?

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint:The dissociation constant is a type of equilibrium constant that measures the propensity when a complex splits into respective molecules, or when a salt compound splits into ions.
-The dissociation constant is denoted as ${{K}_{\alpha }}$ , where$\alpha $ is the degree of dissociation.

Complete step by step answer:
The weak acid dissociates into respective ${{H}^{+}}$ and ${{A}^{-}}$ ions.
So, in the given question, $\alpha $= $0.01$, which is the degree of dissociation of the acid.
${{C}_{1}}=$ $0.1M$; ${{C}_{2}}$ =$0.2M$
Where ${{C}_{1}}$ and ${{C}_{2}}$ are given concentration of acid and the salt.
$T=$$298K$
Where $T$ is the temperature at which the system is present.
We know that the value of dissociation constant is equal to the concentration times square of the degree of dissociation. This equation can be mathematically represented by the following equation,
$\therefore $we know that, ${{K}_{\alpha }}={{C}_{1}}{{\alpha }^{2}}$
Where ${{K}_{\alpha }}$ is the dissociation constant. Now we will substitute the given values into the equation in order to get the value of dissociation constant.
$\Rightarrow $ ${{K}_{\alpha }}$=$0.1\times {{(0.01)}^{2}}$
$\Rightarrow $${{K}_{\alpha }}$=${{10}^{-5}}$
So, after calculation the value of dissociation constant turned out to be ${{10}^{-5}}$.
Now, when the salt is added to the solution, the new $pH$can be calculated using the following equation:
$pH=p{{K}_{\alpha }}+\log (\dfrac{salt}{acid})$
We know that the salt is a sodium salt and the acid is a weak acid. And the value of $p{{K}_{a}}$ is a negative log of the value of the dissociation constant which we just calculated in the previous step.
 $\Rightarrow $ $p{{K}_{\alpha }}=-\log ({{K}_{_{\alpha }}})$
$\Rightarrow $$p{{K}_{\alpha }}=5{{\log }_{10}}10$ =$5$
We know that the value of $\log 10$ to the base $10$ is one, so the value of $p{{K}_{a}}$ turned out to be five.
Now we will calculate the value of $x$
$\Rightarrow $${{10}^{-x}}=\dfrac{5.301}{530}$ =${{10}^{-3}}$
$\Rightarrow $$x=3$
Hence, the value of $x$ is $3$.
Thus, the correct answer is option (A).

Note:
If the value of dissociation constant is more, the greater the strength of the acid or base (in this question, we are talking about acid).
-The weaker the acid, the more difficult it would be for it to dissociate into its constituent ions.
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