Question

a. What is the pH of a solution obtained from adding 184 ml 0.57 M NaOH to 42 mL 0.43 M Phosphoric acid, H₃PO₄?

b.Al(OH)₃is an amphoteric hydroxide

This means it can partially neutralize both acid and base. If aluminum hydroxide is added to a basic solution with a pH of 9.01, what is the new pH after equilibrium is established? There are two equations to consider, but only one is useful for the calculation:

Al(OH)₃(s) ? Al³⁺(aq) + 3OH^-(aq) K_sp = 2.0 x10^-32 M⁴
Al(OH)₃(s) + OH^-(aq) ? Al(OH)₄^-(aq) K = 40

Answer 1

a) Phosphoric acid is a triprotic acid, After dissociation first proton the second proton dissociation is ignorable because of amphoteric nature of H₂PO₄^- . So, for pH calculation we can consider the first proton dissociation.

Hence-

The number of mols of NaOH = (0.187 L)(0.57 mol/L) = 0.10659 mol

The number of mols of H₃PO₄ = (0.042 L)(0.43 mol/L) = 0.01806

According to equation -

          H₃PO₄ (aq) + NaOH (aq) <---------> NaH₂PO₄ (aq) + H₂O (l)

         0.01806         0.10659                           0                   -

           - 0.01806        -0.01806                     +0.01806          -

              0                0.08853                       0.01806

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Thus - pH = pKa + log [NaH₂PO₄ ] / [H₃PO₄]

                = 2.148 + log (0.01806 / 0.08853)

                = 1.457

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b)