Question

In: Chemistry

A 75.0 −mL sample of 1.56×10−2 M Na2SO4(aq) is added to 75.0 mL of 1.28×10−2 M...

A 75.0 −mL sample of 1.56×10−2 M Na2SO4(aq) is added to 75.0 mL of 1.28×10−2 M Ca(NO3)2(aq).what percentage of ca^2+ remains unprecipitate?

Solutions

Expert Solution

[Ca2+] = 0.075 L x (0.0128 mol/L) x (1/0.15 L) = 6.4 x 10-3 M

[SO42-] = 0.075 L x (0.0156 mol/L) x (1/0.15 L) = 7.8 x 10-3 M

                            Ca2+(aq) +   SO42-(aq)                 CaSO4(s)

Initial.................. (6.4 x 10-3 M)........................................ (7.8 x 10-3 M)

Final.................... 0.......................................................... (1.4 x 10-3 M)

CaSO4(s)                                     Ca2+(aq)      +        SO42-(aq)

Initial.................................................. 0.........................(1.4 x 10-3 M)

At equillibrium.................................... x ........................(1.4 x 10-3 M) + x

Ksp = [Ca2+][SO42-]

9.1 x 10-6 = (x) x ((1.4 x 10-3 M) + x)

x2 + (1.4 x 10-3)x - (9.1 x 10-6) = 0

On solving, x = 2.39 x 10-3 M

Therefore, percentage of Ca2+ remains unprecipitate = [(2.39 x 10-3)/ (6.4 x 10-3)] x 100 = 37.34 %


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