Determine the hydronium ion concentration and the pH of a solution that is 0.10

Question

Determine the hydronium ion concentration and the pH of a solution that is 0.10 M NaH_2PO_4 and 0.15 M Na_2 HPO_4 (the K_a for H_2PO_4^- is 6.2 Times 10^-8). Include a chemical equation showing the relevant equilibrium Finish the chemical equation showing the solubility equilibrium for calcium phosphate. Calculate the molar solubility of Ca_3(PO_4)_2 in a 0.20 M Na_3PO_4 solution. As part of your answer, fill in the grid with the relevant numbers or expressions. (K_sp, for calcium phosphate is 1.3 Times 10^-32) How would the solubility of calcium phosphate be affected by the addition of nitric acid solution(increase, decrease or stay the same)? Explain your answer.

Explanation / Answer

1) the mixture of NaH2PO4 and Na2HPO4 will form a buffer solution

The pH of solution will be given as

pH = pKa + log [salt] / [acid]

Ka = 6.2 X 10^-8

pKa = -logKa = 7.21

[Na2HPO4] = [Salt] = 0.15

[NaH2PO4] = [acid] = 0.10

pH = 7.21 + log [0.15/0.10] = 7.39

[H+] = antilog (-pH) = 4.07 X 10^-8

Equilibrium will be

Na2HPO4 + H2O --> NaH2PO4 + NaOH

2)                Ca3(PO4)2 ---> 3Ca+2 + 2PO4-2

Initial             a                   0              0

Change          -x                 +3x           +2x

Equilibrium     a-x                  3x          2x

Ksp = 1.3 X 10^-32 = [3x]^3 [2x]^2 = 108x^5

x = 1.644 X 10^-7 [In water]

Solubility in Na3PO4 will be changed as the concentration of PO4-3 will change due to Na3PO4

[PO4-3] = 0.2

1.3 X 10^-32 = [3x]^3 [2x+0.2]^2

2x << 0.2

1.3 X 10^-32 = [3x]^3 X 0.2 X 0.2 = 1.08x^3

x = 2.29 X 10^-11

so solubility will decrease in NaF as compared to water

3) There will be no change as HNO3 will react with calcium ion to form calcium nitrate which is again soluble in water

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