Question

. You have 45 mL of a 0.84 M solution of HCOOH (K 1.8 x 10-4). You titrate it with 0.51 M NaOH What is the pH at the half equivalence point (the point where you have added half the number of moles of base as you had acid)? (a) 2.08 (b) 2.51 (c) 2.77 (d) 3.30 (e) 3.74

Answer 1

TO NOTE THAT AT HALF EQUIVALENCE POINT THE CONCENTRATION OF ANION FORMED IS EQUAL TO CONCENTRATION OF THE REMAINING ACID.

HCOOH + NaOH $\Leftrightarrow$ H₂O +HCOO^-Na⁺

AT HALF EQUIVALENCE POINT [ HCOOH ] = [HCOO^-Na⁺]

AT HALF EQUIVALENCE POINT HALF OF THE VOLUME OF SODIUM HYDROXIDE NEUTRALISES HALF OF THE ACID HCOOH, SO CONCENTRATION SALT FORMED = CONCENTRATION OF ACID REMAINING.

IN THE HENDERON EQUATION , THEN pH BECOMES EQUAL TO pKa VALUE OF HCOOH

SO pH = pKa

= -LOG(1.8X 10^-4)

= 4- LOG1.8

= 4-0.25527

= 3.74

OPTION e IS MATCHING