Question

4 1, 2, 3, and 4 moles are taken of $H_3PO_4$, $NaH_2PO_4$, $Na_2HPO_4$, $Na_3PO_4$ respectively with $ka_1 10^{-3}$, $ka_2 10^{-7}$, and $ka_3 10^{-13}$ find the resulting PH when these are mixed to gether to form aqueous 1L soln.

Answer 1

pH ≈ 6.8

Answer 2

Solution

We start with a mixture of 10 moles of phosphate (P) keeping track of the number of “acidic protons.” In the forms given the number of protons bound per mole is:

• $H_3PO_4$: 3 protons × 4 moles = 12
• $NaH_2PO_4$ (as $H_2PO_4^−$): 2 protons × 1 mole = 2
• $Na_2HPO_4$ (as $HPO_4^{2−}$): 1 proton × 2 moles = 2
• $Na_3PO_4$ (as $PO_4^{3−}$): 0 protons × 3 moles = 0

Total “proton‐units” = 12 + 2 + 2 = 16
Total phosphate moles = 4 + 1 + 2 + 3 = 10

Thus the average number of protons per P is 16/10 = 1.6.

Note that if the system is allowed to equilibrate by internal proton transfer (with water merely mediating the exchange) the final speciation must conserve both phosphorus and the total H–content on the phosphate, but rearranged so that each phosphate is “1.6‐protonated” on average.

A natural way to obtain an average of 1.6 is to have only the second dissociation forms present, namely $H_2PO_4^−$ (which has 2 protons) and $HPO_4^{2−}$ (which has 1 proton). Suppose at equilibrium we have:

Let moles of $H_2PO_4^−$ = x and moles of $HPO_4^{2−}$ = 10 – x.

The overall proton count would be: 2x + 1(10 – x) = x + 10.

Setting this equal to 16 gives:

x + 10 = 16 ⟹ x = 6.

Thus the equilibrium mixture is:

6 moles $H_2PO_4^−$ and 4 moles $HPO_4^{2−}$.

Now the equilibrium controlling the interconversion is the second dissociation:

$H_2PO_4^− \rightleftharpoons H^+ + HPO_4^{2−}$ with $K_2 = 10^{−7}$.

In a 1 L solution the concentrations are:

$[H_2PO_4^−]$ = 6 M and $[HPO_4^{2−}]$ = 4 M.

Writing the equilibrium expression:

$K_2 = \frac{([H^+][HPO_4^{2−}])}{[H_2PO_4^−]}$

⟹ $10^{−7} = [H^+] \cdot \frac{4}{6} = [H^+] \cdot \frac{2}{3}$

Solve for $[H^+]$:

$[H^+] = 10^{−7} \times \frac{3}{2} = 1.5 \times 10^{−7}$ M.

Thus the pH is:

pH = –log($1.5 \times 10^{−7}$) ≈ 6.82.

Final Answer: pH ≈ 6.8

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Summary for JEE/NEET:

• Core Idea: The initial mix has a net proton content giving an average protonation of 1.6 per phosphate. In equilibrium, all phosphate is redistributed into only $H_2PO_4^−$ (2 H’s) and $HPO_4^{2−}$ (1 H) in a 6 : 4 ratio. Then using the Henderson–Hasselbalch relation (or the equilibrium constant for the second dissociation) we obtain the pH.
• Answer: pH ≈ 6.8