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Question: A \(0.1M\) solution of weak acid \(HA\) is \(1%\) dissociated at \(298K\). The new \(pH\) when \(0.2...

A 0.1M0.1M solution of weak acid HAHA is 11% dissociated at 298K298K. The new pHpH when 0.2M0.2M of NaANaA is added to it is 530×10x530\times {{10}^{-x}}, then what will be the value of xx?

Explanation

Solution

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 Kα{{K}_{\alpha }} , whereα\alpha is the degree of dissociation.

Complete step by step answer:
The weak acid dissociates into respective H+{{H}^{+}} and A{{A}^{-}} ions.
So, in the given question, α\alpha = 0.010.01, which is the degree of dissociation of the acid.
C1={{C}_{1}}= 0.1M0.1M; C2{{C}_{2}} =0.2M0.2M
Where C1{{C}_{1}} and C2{{C}_{2}} are given concentration of acid and the salt.
T=T= 298K298K
Where TT 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, Kα=C1α2{{K}_{\alpha }}={{C}_{1}}{{\alpha }^{2}}
Where Kα{{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 Kα{{K}_{\alpha }}=0.1×(0.01)20.1\times {{(0.01)}^{2}}
\Rightarrow Kα{{K}_{\alpha }}=105{{10}^{-5}}
So, after calculation the value of dissociation constant turned out to be 105{{10}^{-5}}.
Now, when the salt is added to the solution, the new pHpHcan be calculated using the following equation:
pH=pKα+log(saltacid)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 pKap{{K}_{a}} is a negative log of the value of the dissociation constant which we just calculated in the previous step.
\Rightarrow pKα=log(Kα)p{{K}_{\alpha }}=-\log ({{K}_{_{\alpha }}})
\Rightarrow pKα=5log1010p{{K}_{\alpha }}=5{{\log }_{10}}10 =55
We know that the value of log10\log 10 to the base 1010 is one, so the value of pKap{{K}_{a}} turned out to be five.
Now we will calculate the value of xx
\Rightarrow 10x=5.301530{{10}^{-x}}=\dfrac{5.301}{530} =103{{10}^{-3}}
\Rightarrow x=3x=3
Hence, the value of xx is 33.
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.