Question

20 ml solution containing 0.1 m na2co3, 0.15 m naoh and 0.20 m nahco3 was titrated with 25 ml of a solution of hcl using phenolphthalein indicator. Hence strength of hcl solution was

Answer 1

Answer:

Explanation:

HENCE PHENOLPHTHALEIN IS USED

NAHCO3 REMAINS UN IONISED

SO EQUIVALENT OF NA2CO3+ EQUIVALENT OF NAOH = 25X

M*V*F+M*V*F=25X

0.1(20-V)+0.15(V) = 25X

2+0.05V=25X

2+0.05V / 25 =X

EQUIVALENT OF HCL = (2+0.05V)

WHERE V=VOLUME OF NAOH