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?

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Expert Solution

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

[SO₄^2-] = 0.075 L x (0.0156 mol/L) x (1/0.15 L) = 7.8 x 10^-3 M

Ca²⁺(aq) + SO₄^2-(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) + SO₄^2-(aq)

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

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

Ksp = [Ca²⁺][SO₄^2-]

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 %

queen_honey_blossom answered 1 year ago

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