Chemistry, asked by yashusharma96, 11 months ago

Ka for an Acid (HA) is5×10^-6 .
What is the pH of 0.2 M solution of HA.
Find also the molar concentration of À​

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Answered by tarunjala06
0

Answer:

Secondary School Chemistry 5 points

If the Ka of a weak monoprotic acid is 4* 10-6 and

its concentration is 0.2 M then pH of solution is

(1) 305

(2) 3.72

(3) 3.99

(4)4.62

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shubham0204

shubham0204 Expert

Answer:

See below.

Explanation:

We are given,

\begin{aligned}K_{a}=4\times 10^{-6}\\ C=0.2\end{aligned}

For a weak acid, we know,

\begin{aligned}K_{a}=C\alpha ^{2}\\ \Rightarrow \alpha =\left( \dfrac {K_{a}}{c}\right) ^{\dfrac {1}{2}}\end{aligned}

Calculating the degree of dissociation ( alpha ),

\begin{aligned}\alpha ^{2}=\dfrac {4\times 10^{-6}}{0.2}=2\times 10^{-5}\\ \Rightarrow \alpha =0.447\times 10^{-2}\end{aligned}

Calculating the concentration of H+ ions,

\begin{aligned}\left[ H^{+}\right] =C\alpha =0.2\times 0.477\times 10^{-2}\\ =0.0954\times 10^{-2}\\ =9.54\times 10^{-4}\end{aligned}

Calculating pH,

\begin{aligned}pH=-\log \left[ H^{+}\right] \\ =-\log \left( 9.54\times 10^{-4}\right) \\ =-\left( 0.98-4\right) \\ =3.02\end{aligned}

The correct option should be 3.05 ( nearest answer ).

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