Answer:
0.200M H₃PO₄
0.600N H₃PO₄
pH = 1.46
Explanation:
The acid-base reaction of phosphoric acid (H₃PO₄) with LiOH is:
3 LiOH + H₃PO₄ → Li₃PO₄ + 3H₂O
<em>Where 3 moles of LiOH reacts per mole of H₃PO₄</em>
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Moles of LiOH are:
0.030L× (0.5mol / L) = 0.0150 moles of LiOH
Moles of acid neutralized 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 =<em> 0.200M H₃PO₄</em>
Normality is:
0.200M × (3N H⁺ / 1M H₃PO₄) = <em>0.600N H₃PO₄</em>
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 in Ka formula:
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
As, [H⁺] = 0.0343M
pH = - log [H⁺],
<em>pH = 1.46</em>
