Question

To a 25.00 mL volumetric flask, a lab technician adds a 0.225 g sample of a weak monoprotic acid, HA , and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0849 M KOH . She reaches the endpoint after adding 42.35 mL of the KOH solution. Determine the number of moles of the weak acid in the solution.

Answer 1

The number of moles of the weak acid in the solution is 0.00359

Calculation:

1. The number of mole of KOH in the solution:

Given:

Molarity of KOH = 0.0849 M

Volume = 42.35 mL = 42.35 / 1000 = 0.04235 L

Mole of KOH = Mole = Molarity x Volume

                      = 0.0849 ×  0.04235

                      = 0.00359 mole

2. The number of mole of a weak acid, HA:  

Balanced equation:

HA + KOH → KA + H₂O

From the above equation, A mole of HA and a mole of KOH react together. Therefore,

0.00359 moles of HA will also react with 0.00359 moles of KOH.

0.00359  moles of base use = 1/1 x 0.00359 = 0.00359 moles of HA

Thus, there are 0.00359 moles of a weak acid in the solution.

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