Question #191511

For a titration of 20 mL ofÂ 0.80M formic acidÂ (K_{a}Â = 3.7 x 10^{-4}, values are made up), with NaOH (0.80M), the pH of the solutions at half Equivalence point and atÂ the EquivalenceÂ point are:

Expert's answer

HALF EQUIVALENCE POINT

At half equivalence point is point at which exactlyÂ **half**Â 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

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