Question

For a titration of 20 mL of 0.80M formic acid (K_a = 3.7 x 10^-4, values are made up), with NaOH (0.80M), the pH of the solutions at half Equivalence point and at the Equivalence point are:

Answer 1

HALF EQUIVALENCE POINT

At half equivalence point is point at which exactly half of the acid in the buffer solution has reacted with our titrent.

That mean...

At half equivelence point

$[HCOOH]=[HCOO^-]$

Thus, we get our P^H=P^ka

therefore P^H at half equivelence point is =4-(0.56)

=3.44

EQUIVALENCE POINT

At equivalence point salt of sodium formate is present,which is a salt of weak acid and strong base....

And its p^H is given by...

$p^H=7+0.5 p^{Ka}+0.5 logC$

Where C is concentration of salt,

As 20 ml 0.80M acid require 20 ml 0.80M base

Therefore C is

${{20\times0.80}\over20+20}={16\over40}=0.4$

p^H =7+0.5 (3.44)+ 0.5(-0.39)

=7+1.72 - .195

=8.525 ......answer