For Ca(OH)₂,K_sp = 4.0 × 10−⁶.What will occur if 1.0 L of 0.100 M Ca(NO₃)₂ is prepared in a solution that is buffered at pH 12.50?

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between Ca(OH)2 and H2S is: Ca(OH)2 + H2S ↔ CaS + 2H2O The solubility product expression for CaS is: K_sp = [Ca2+][S2-] In a buffered solution at pH 12.50, the concentration of hydroxide ions is: [OH-] = 10^(-2.50) M = 3.16×10^(-3) M The concentration of Ca2+ ions from Ca(OH)2 is: [Ca2+] = [OH-] = 3.16×10^(-3) M The concentration of S2- ions from H2S is: [S2-] = (4.0 × 10^(-16))/(0.100) = 4.0×10^(-15) M The ionic product Q for CaS is: Q = [Ca2+][S2-] = (3.16×10^(-3))(4.0×10^(-15)) = 1.264×10^(-17) Since Q > K_sp, a precipitate of CaS will form. Therefore, the correct answer is A.

For Ca(OH)2,Ksp = 4

For Ca(OH)₂,K_sp = 4