30 ml of a 0.5 N lithium hydroxide standard are used to titrate 25 ml of a phosphoric

acid sample. What is the molarity, normality and pH of the acid?

0.200M H₃PO₄

0.600N H₃PO₄

pH = 1.46

Explanation:

The neutralization reaction of phosphoric acid (H₃PO₄) with LiOH is:

3 LiOH + H₃PO₄ → Li₃PO₄ + 3H₂O

Where 3 moles of LiOH reacts per mole of H₃PO₄

Moles of LiOH used in the standarization are:

0.030L * (0.5mol / L) = 0.0150 moles of LiOH

Moles of acid are:

0.0150 moles of LiOH * (1 mole H₃PO₄ / 3 moles LiOH) = 0.005 moles H₃PO₄

As volume of acid was 25mL, molarity is:

0.005mol H₃PO₄ / 0.025L = 0.200M H₃PO₄

Normality is:

0.200M * (3N H⁺ / 1M H₃PO₄) = 0.600N H₃PO₄

H₃PO₄ dissolves in water thus:

H₃PO₄ ⇄ H₂PO₄⁻ + H⁺

Ka = 7.1x10⁻³ = [H₂PO₄⁻] [H⁺] / [H₃PO₄]

Where molar concentrations in equilibrium will be:

[H₂PO₄⁻] = X

[H⁺] = X

[H₃PO₄] = 0.200M - X

Replacing:

7.1x10⁻³ = [X] [X] / [0.200 - X]

1.42x10⁻³ - 7.1x10⁻³X = X²

0 = X² + 7.1x10⁻³X - 1.42x10⁻³

Solving for X:

X = - 0.04M →False answer, there is no negative concentrations.

X = 0.0343M

Thus, [H⁺] = 0.0343M

As pH = - log [H⁺],

pH = 1.46