Question

To a 25.00 mL volumetric flask, a lab technician adds a 0.250 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 0.0846 M KOH . She reaches the endpoint after adding 46.83 mL of the KOH solution. Determine the number of moles of the weak acid in the solution.

Answer 1

Answer: The number of moles of the weak acid in the solution is 0.00396

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$      

a) ${\text{Moles of}KOH=0.0846M\times 0.04683L=0.00396mol$

The balanced chemical reaction is:

$HA+KOH\rightarrow KA+H_2O$  

1 mole of base uses = 1 mole of $HA$  

0.00396  moles of base use = $\frac{1}{1}\times 0.00396=0.00396$ moles of $HA$  

Thus the number of moles of the weak acid in the solution is 0.00396