Question

You titrated 100.00 mL of a 0.025 M solution of benzoic acid (HBz) with 0.100 M NaOH to the equivalence point. Ka = 6.2x10-5

Answer 1

100 ML OF 0.025 M SOLUTION OF BENZOIC ACID =0.3053 GMS OF BENZOIC ACID PRESENT IN 100 ML SOLUTION.,

MOLES OF BENZOIC ACID=0.3053/122.12=0.0025 MOLES

REACTION IS BALANCED , THERE FORE 1 MOL ACID REQUIRES 1 MOL BASE,

MEANS 0.0025 MOLES OF NaOH IS NEEDED TO REACH EQUIVALENCE POINT.

0.0025*40=0.1 GMS, , THERE FORE 25 ML OF 0.1 MOLAR SOLUTION OF NaOH IS REQUIRED.

40 GMS=1000 ML=1 MOLAR, THEN FOR 0.1 M=?

4 GMS IN 1000 ML =0.1 MOLAR

4 GMS=1000 ML

0.1 GMS =?

0.1*1000=100/4=25 ML ML 0.1 MOLAR NaOH