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.
Answers
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.