7. A student titrated a 15.00-mL sample of a solution containing a weak, monoprotic acid with NaOH. The titration required 17.73 mL of 0.1036 M NaOH to reach the equivalence point. Calculate the concentration (in M) of the weak acid in the sample.

The concentration of the weak acid, HA in the sample is 0.123 M

We'll begin by calculating the number of mole of NaOH in the solution. This can be obtained as follow:

Molarity of NaOH = 0.1036 M

Volume = 17.73 mL = 17.73 / 1000 = 0.01773 L

Mole = Molarity x Volume

Mole of NaOH = 0.1036 × 0.01773

HA + NaOH —> NaA + H₂O

From the balanced equation above,

1 mole of HA reacted with 1 mole of NaOH

Therefore,

0.00184 mole of HA will also react with 0.00184 mole of NaOH.

Mole of HA = 0.00184 mole

Volume = 15 mL = 15 / 1000 = 0.015 L

Concentration = mole / Volume

Concentration of HA = 0.00184 / 0.015

Therefore, the concentration of the weak acid, HA in the sample is 0.123 M

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 The concentration of the weak acid in the sample is 0.1224 M

Explanation:

The balanced chemical reaction is :

To calculate the volume of acid, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

Thus the concentration (in M) of the weak acid in the sample is 0.1224

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