Question

A 25-mL solution of 0.25 M HF (Ka =7.1 x 10-4) is titrated with a 0.15 M solution of NaOH.

Determine the pH of the solution at the equivalence point. Select one: a. 5.94 b. 10.8 c. 8.90 d. 8.60 e. 8.06

Answer 1

The volume of NaOH required to titrate HF is given By

M1V1 = M2V2

Where,

M1 = Concentration of HF = 0.25 M

V1 = Volume of HF = 25 mL

M2 = Concentration of NaOH =  0.15 M

V2 = Volume of NaOH = ? (to be found)

V - M/V M2 0.25 x 25 1 0.15 ..V2 = = 41.66 mL

Volume of NaOH required at equivalence point = 41.66 mL

At equivalence point, moles of acid = moles of base

Moles of HF at equivalence point = concentration x volume (in L) = 0.25 x (25/1000 ) = 0.00625 Moles

Therefore, No .of moles of NaOH at equivalence point = 0.00625 Moles

Reaction:  

$HF + NaOH \rightarrow NaF + H_2O$

So,

No.of moles of NaF formed at equivalence point = no.of moles of HF = no .of moles of NaOH = 0.00625 moles

This NaF formed is a weak base and gets hydrolysed to give OH- ions

Concentration of NaF at equivalence point

$[NaF]=\frac{no.\;\;of\;\; moles}{total \;Volume (in\;L)}$

0.00625 NaF] = (41.66 +257 X 1000 = 0.094 M

Now, NaF hydrolyses and follows the equillibrium

$NaF + H_2O \rightleftharpoons HF + Na^+ + OH^-$

Kb = Kw/Ka  = 10-14 / (7.1 x 10-4) = 1.4 x 10-11

ICE for this equillibriu

	[NaF]	[HF]	[OH-]
Initial	0.094	0	0
Change	-x	+x	+x
Equilibrium	0.094-x	+x	+x

$K_b = \frac{[HF][OH^-]}{[NaF]} = \frac{x^2}{0.094-x}$

But,

Kb = 1.4 x 10-11   So equating above relation with this

But 0.094 >>>x (since NaF is weak base), therefore 0.094 - x $\approx$ 0.094 M

so, the relation becomes

$\frac{x^2}{0.094} = 1.4 \times 10^{-11}\\ \therefore x = \sqrt{1.4 \times 0.094 \times 10^{-11}} = 1.14 \times 10^{-6}$

so, [OH-] = 1.14 x 10-6 M

pOH = -log [OH-] = -log( 1.14 x 10-6 ) = 5.94

But, pH = 14 - pOH

Therefore pH = 14 - 5.94 = 8.06

Therefore pH at equivalence point is e) 8.06

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