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Mole Concept

A student use...

Question

A student used an acid-base titration to determine the molar mass of solid weak monoprotic acid like $K H P$ . This student titrated a weighed sample of the solid acid, after dissolving it in water in a flask, with a standardized $N a O H$ solution in a buret.
Which of the following could explain why the student obtained a molar mass that was too large?

Some of the solid monoprotic acid still remained on the weighing paper and was not transferred to the flask for titration.

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More water was used to dissolve the solid monoprotic acid than what the lab called for.

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Addition of some base beyond the endpoint of the titration.

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The burest was not conditioned with the standardized

N a O H

solution, but with distilled water at the beginning of the titration.

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Solution

The correct option is A Some of the solid monoprotic acid still remained on the weighing paper and was not transferred to the flask for titration.
Let $w$ be weight of $K H P$ taken and $W$ be the Molar mass of $K H P$ .
At equivalence point, $M o l e s$ of $K H P = M o l e s$ of $N a O H$
Let, $M$ and $V$ be molarity and volume of $N a O H$
$∴ \frac{w}{W} = M \times V$
$∴ W = \frac{w}{M \times V}$
Now, $W$ can be large when $w$ is less than what was excepted due to which $V$ is also less, since volume $V$ required for less $w$ will be less.
This might happen if some $K H P$ was left on righting power.

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