Only thoruoghly handwritten solutions are credited. thanks A closed system solut

Question

Only thoruoghly handwritten solutions are credited. thanks

A closed system solution contains the solid calcite (CaCO_3). You decide to add 0.001 M of carbonic acid (H_2CO_3). At equilibrium, the pH is 9. 4 and the concentration of H_2CO_3 is equal to 10^-6.68 M. Using the information provided below, what is the concentration of calcium (Ca^2+) dissolved at equilibrium? The (=) sign represents equilibrium. CaCO_3_(s) = Ca^2+_(aq) + CO_3^2- (aq); K_sp = 10^-8.48 H_2CO_3 + H_2O = H_3O^+ + HCO_3^-: K_a, 1 = 10^-6.33 HCO_3^- + H_2O = H_3O^+ + CO_3^2-; K_a, 2 = 10^-10.33

Explanation / Answer

CaCO3------> Ca2+ + CO32-

let x be the molar solubility of CaCO3. The addition of 0.001M Carbonic acid, the concentraion at equlbirum are

Ca2+ = x mol/L

CO32- = 0.001+x mol/L

Ksp = [Ca2+][CO32-]= x (0.001+x)

10-8.48 = 0.001x + x2

Neglecting x2 as it is very samll compared to 0.001x

x = 10-8.48 / 0.001 = 1.11 x 10-6 M

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