Question

Suppose you have 50.4 mL of a solution of H₃PO₄. You titrate it with a 0.10 M solution of LiOH and use 15.2 mL to reach the endpoint.

A.) What is the concentration of H₃PO₄?

B.) What was the pH of the original H₃PO₄ solution?

Answer 1

A)

Consider reaction, H₃PO₄ (aq) + 3 LiOH (aq) Li₃PO₄ (aq) + 3 H₂O (l)

From reaction, Stoichiometric ratio = no. of moles of acid / no. of moles of base = 1 / 3

We have correlation, M _acid V _acid = M _base V _base Stoichiometric ratio

M _acid = M _base V _base Stoichiometric ratio /  V _acid

= 0.10 M 15.2 ml ( 1/3 ) / 50.4 ml

= 0.01005 M

ANSWER : [ H₃PO₄ ] = 0.01005 M

B)

Consider reaction, H₃PO₄ (aq) + H₂O (l) H₃O ⁺ (aq) + H₂PO₄^- (aq)

Equilibrium constant for above reaction is , K a = [ H₃O ⁺ ] [ H₂PO₄^- ] / [ H₃PO₄ ] = 7.11 10 ^-03 .

Let's use ICE table to find out equilibrium concentrations of  H₃O ⁺, H₂PO₄^- and H₃PO₄.

Concentration ( M )	H3PO4	H3O +	H2PO4-
I	0.01005
C	- X	+ X	+X
E	0.01005 - X	X	X

Substituting these values in K a expression, we get

K a = ( X ) ( X ) / 0.01005 - X = 7.11 10 ^-03 .

X ² / 0.01005 - X = 7.11 10 ^-03 .

X ² = 0.01005 - X ( 7.11 10 ^-03 )

X ² = 7.146 10 ^-05 - 7.11 10 ^-03 X

X ² + 7.11 10 ^-03 X - 7.146 10 ^-05 = 0

Comparing to a X 2 + b X + c = 0 , we get a = 1 , b = 7.11 10 ^-03 and c = - 7.146 10 ^-05 .

Now, solve for X.

X = - b +/- b ² -4 ac / 2 a

X = - b +/- ( 7.11 10 ^-03 ) ² - 4 ( 1) ( - 7.146 10 ^-05 ) / 2 (1)

X = - b +/- 3.3637 10 ^-04 / 2

X = - b +/- 0.01834 / 2

X = -  7.11 10 ^-03 + 0.01834 /2 = 0.005615

or

X = -  7.11 10 ^-03 - 0.01834 /2 = - 0.012725

Acceptable value of X is 0.005615.

[ H₃O ⁺ ] = [ H₂PO₄^- ] = X = 0.005615 M

We have relation, pH = -log [ H₃O ⁺ ]

pH = - log ( 0.005615 ) = 2.25

ANSWER : pH of H₃PO₄ solution is 2.25