Question

When Benzoic acid C6H5COOH dissolved in 75 mL of water is titrated with NaOH,  the equivalence point is reached with 31.78 mL of 0.09251 M NaOH solution. What is the concentration of benzoic acid solution.
How much mole of benzoic acid is present in the solution.​

Answer 1

Answer:

I The reaction that is taking place is as follows:

C6H5COOH + NaOH--------> C6H5COONa + H-OH

Molar concentration of Benzoic acid=?

volume of Benzoic acid solution= 75 ml =>75/1000 litres =>3/40 litres

The reaction completes when 31.78 ml of 0.09251 M NaOH solution has been added.

molarity=(moles of solute)/litres of solution

0.09251 M = x mol/ (31.78/1000) litres

x=0.09251  *( 31.78/1000)

x=0.00294 mol of NaOH

According to the equation, 1 part of Benzoic acid reacts with 1 part of NaOH.

1 part=0.00294 mol

Hence, 0.00294 mol of Benzoic acid is used in the reaction

molarity of benzoic acid solution=(moles of benzoic acid)/volume of solution

=>0.00294/(3/40)

=>0.00294*40/3

=>0.0392 M

Hence, the molar concentration of Benzoic acid solution is 0.0392 M and the number of moles of C6H5COOH present in solution are 0.00294 mol.