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Question
50 ml of a weak monoprotic acid was titrated with 0.1 M NaOH solution. If the pH of solution after adding 50 ml and 75 ml of base is 4.699 and 5 then initial concentration of weak acid was:
(take log 2 = 0.301)
(take log 2 = 0.301)
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Solution
The correct option is C 0.3 M
We know,
WA+NaOH→Salt +H2O
where, WA = weak acid
The final pH is below 7, hence acidic buffer will form.
Thus,
pH=pKa+log[salt][acid]
(pH)1=pKa+log0.1×50C×50−0.1×50 ....(1);
Since monoprotic acid, we can take the volume of the acid also be 50 mL
(pH)2=pKa+log0.1×75C×50−0.1×75 ....(2)
Now, (2)-(1) and taking antilog,
⇒2=32(C×50−5C×50−7.5)
⇒C=0.3 M
We know,
WA+NaOH→Salt +H2O
where, WA = weak acid
The final pH is below 7, hence acidic buffer will form.
Thus,
pH=pKa+log[salt][acid]
(pH)1=pKa+log0.1×50C×50−0.1×50 ....(1);
Since monoprotic acid, we can take the volume of the acid also be 50 mL
(pH)2=pKa+log0.1×75C×50−0.1×75 ....(2)
Now, (2)-(1) and taking antilog,
⇒2=32(C×50−5C×50−7.5)
⇒C=0.3 M
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