Question

25.0 mL of 0.150 M Na3PO4 are combined with 25.0 mL of 0.200 M Sr(NO3)2. Calculate the concentrations of all ionic species (except H+ and OH-) at equilibrium. Ksp( Sr3(PO4)2 ) = 1.0x10^-31.

Answer: Na+ = 0.225 M

PO4= 0.0084 M

Sr= 1.12 x 10^-9 M

NO3 = 0.220 M

Answer 1

After mixing the solution

[Na3PO4] = 0.150M/(50ml/25ml) = 0.075M

[Na⁺] = 3× 0.075M = 0225M

[Sr(NO3)2] = 0.200M/(50ml/25ml) = 0.100M

[NO3^-] = 2×0.100M = 0.200M

Consider the complete reaction of formation of Sr3(PO4)2

3Sr²⁺(aq) + 2PO4^3-(aq) ------> Sr3(PO4)2(s)

Stoichiometrically, 3 moles of Sr²⁺ react with 2moles of PO4^3-

[Sr²⁺] = 0.200M/2 = 0.100M

[PO4^3-] = 0.150M/2 = 0.075M

0.100M Sr²⁺ react with 0.0667M PO4^3-

After reaction

[Sr²⁺] = 0

[PO4^3-] = 0.075M - 0.0667 M = 0.0083

Now ,consider the solubility equillibrium of Sr3(PO4)3

Sr3(PO4)3(s) <------> 3Sr²⁺(aq) + 2PO4^3-(aq)

Ksp = [Sr²⁺]³[PO4^3- ]² = 1.0×10^-31

Initial concentration

[Sr²⁺] =0

[PO4^3-] =0.0083

change in concentration

[Sr²⁺] = +x

[PO4^{3-] = +x}

Equillibrium concentration

[Sr²⁺] = x

[PO4^3- ] = 0.0083 + x

so,

x³(0.0083 +x)² = 1.0 ×10^-31

x = 1.13×10^-9

Therefore, at equillibrium

[Sr²⁺] = 1.13×10^-9M

[PO4^3-] = 0.0083M

[Na⁺] = 0.225M

[NO3^-] = 0.200M