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Lewis Theory

During the ti...

Question

During the titration of a weak diprotic acid $(H_{2} A)$ against a strong base $(N a O H)$ , the $p H$ of the solution half-way to the first equivalent point and that at the first equivalent point are given respectively by :

A

p K_{a_{1}}

and

p K_{a_{1}} + p K_{a_{2}}

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B

\sqrt{K_{a_{1}} C}

and

\frac{p K_{a_{1}} + p K_{a_{2}}}{2}

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C

p K_{a_{1}}

and

\frac{p K_{a_{1}} + p K_{a_{2}}}{2}

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D

p K_{a_{1}}

and

p K_{a_{2}}

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Solution

The correct option is C $p K_{a_{1}}$ and $\frac{p K_{a_{1}} + p K_{a_{2}}}{2}$
During midway of first equivalent point, buffer is formed and the concentration of salt and acid is 1. So, using the formula of buffer, pH = pKa + 0 = pKa.
During first equivalent point, pH is found by using by using the average values of first ionisation and second constant.
The final equivalence point is attained by adding another $10 m l$ ,or a total of $30 m l$ of the titrant to the weak polyprotice acid. image created by Heather Yee.

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