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Gramar

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edited Mar 8 at 21:25

I know the formulae for weak acid-weak base salt of AB type. A peculiar question made me ask this. Do the formula for derived for AB type also hold for A3B type sal. For example: This is the question, I got from my teacher

My attempt at solution:

Equation-1:$$\ce{3NH^+_4 +PO_4^{-3} + 3H_2O\rightleftharpoons 3NH_4OH + H_3PO_4}$$

$$K_h=\ce{\cfrac{[NH_4OH]^3[H_3PO_4]}{[NH_4^+]^3[PO_4^{-3}]}}=\cfrac{[3ch]^3×[ch]}{[3c(1-h)]^3[c(1-h)]}$$ $$\implies K_h=\frac{27c^4h^4}{27c^4(1-h)^4}\approx h^4$$ Equation-2: $$\ce{NH_4OH\rightleftharpoons NH_4^+ + OH^-}$$ $$K_b=\ce{\cfrac{[NH_4^+][OH^-]}{[NH_4OH]}}$$ Equation-3: $$\ce{H_3PO_4\rightleftharpoons 3H^+ + PO_4^{-3}}$$ $$K_a=\ce{\cfrac{[H^+]^3[PO_4^{-3}]}{[H_3PO_4]}}$$ Equation-4: $$K_w=[H^+][OH^-]$$

From the above equations, $$K_h×K_a×K_b^3=K_w^3$$$$\implies h^4=\cfrac{K^3_w}{K^3_b×K_a}$$

$$[H^+]^3=\cfrac{K_ach}{c-ch}\approx K_ah\implies [H^+]=\left(\cfrac{K_wK_a}{K_b}\right)^{(1/4)}$$ $$pH=\frac{1}{4}\left(pK_w+pK_a-pK_b\right)$$

This gives half of the correct answer.

Teacher's Solution

The formula given here is applicable to all scenarios so use $$pH=\frac{1}{2}\left(pK_w+pK_a-pK_b\right).$$ This gives $7.24$ which is the correct answer.

My question is :

Is the formula for AB$AB$ type weak acid-weak base also applicable to A_3B $A_3B$ type weak acid-weak base salt? What is the correct formula for A_3B$A_3B$ salts?
How to derive pH of A_xB_y$ A_xB_y$ type weak acid-weak base salt??

My attempt at solution:

Equation-1:$$\ce{3NH^+_4 +PO_4^{-3} + 3H_2O\rightleftharpoons 3NH_4OH + H_3PO_4}$$

From the above equations, $$K_h×K_a×K_b^3=K_w^3$$$$\implies h^4=\cfrac{K^3_w}{K^3_b×K_a}$$

$$[H^+]^3=\cfrac{K_ach}{c-ch}\approx K_ah\implies [H^+]=\left(\cfrac{K_wK_a}{K_b}\right)^{(1/4)}$$ $$pH=\frac{1}{4}\left(pK_w+pK_a-pK_b\right)$$

This gives half of the correct answer.

Teacher's Solution

The formula given here is applicable to all scenarios so use $$pH=\frac{1}{2}\left(pK_w+pK_a-pK_b\right).$$ This gives $7.24$ which is the correct answer.

My question is :

Is the formula for AB type weak acid-weak base also applicable to A_3B type weak acid-weak base salt? What is the correct formula for A_3B salts?
How to derive pH of A_xB_y type weak acid-weak base salt??

My attempt at solution:

Equation-1:$$\ce{3NH^+_4 +PO_4^{-3} + 3H_2O\rightleftharpoons 3NH_4OH + H_3PO_4}$$

From the above equations, $$K_h×K_a×K_b^3=K_w^3$$$$\implies h^4=\cfrac{K^3_w}{K^3_b×K_a}$$

$$[H^+]^3=\cfrac{K_ach}{c-ch}\approx K_ah\implies [H^+]=\left(\cfrac{K_wK_a}{K_b}\right)^{(1/4)}$$ $$pH=\frac{1}{4}\left(pK_w+pK_a-pK_b\right)$$

This gives half of the correct answer.

Teacher's Solution

The formula given here is applicable to all scenarios so use $$pH=\frac{1}{2}\left(pK_w+pK_a-pK_b\right).$$ This gives $7.24$ which is the correct answer.

My question is :

Is the formula for $AB$ type weak acid-weak base also applicable to $A_3B$ type weak acid-weak base salt? What is the correct formula for $A_3B$ salts?
How to derive pH of $ A_xB_y$ type weak acid-weak base salt??

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asked Mar 8 at 21:19

Aurelius

Hydrolysis of $A_3B$ type weak acid-weak base salt

My attempt at solution:

Equation-1:$$\ce{3NH^+_4 +PO_4^{-3} + 3H_2O\rightleftharpoons 3NH_4OH + H_3PO_4}$$

From the above equations, $$K_h×K_a×K_b^3=K_w^3$$$$\implies h^4=\cfrac{K^3_w}{K^3_b×K_a}$$

$$[H^+]^3=\cfrac{K_ach}{c-ch}\approx K_ah\implies [H^+]=\left(\cfrac{K_wK_a}{K_b}\right)^{(1/4)}$$ $$pH=\frac{1}{4}\left(pK_w+pK_a-pK_b\right)$$

This gives half of the correct answer.

Teacher's Solution

The formula given here is applicable to all scenarios so use $$pH=\frac{1}{2}\left(pK_w+pK_a-pK_b\right).$$ This gives $7.24$ which is the correct answer.

My question is :