A lab technician adds 0.015 mol of KOH to 1.00 L of 0.0010 M Ca(NO_3)2 K_sp = 6.

Question

A lab technician adds 0.015 mol of KOH to 1.00 L of 0.0010 M Ca(NO_3)2 K_sp = 6.5 Times 10^-6 for Ca(OH)_2). Which of the following statements is correct? A The presence of KOH will raise the solubility of Ca(NO_3)2. B. The solution is unsaturated and no precipitate forms. C. One must know K_sp for calcium nitrate to make meaningful predictions on this system. D. Calcium hydroxide precipitates until the solution is saturated. E. The concentration of calcium ions is reduced by the addition of the hydroxide ions. You have a saturated solution of Ag_2SO_4. It follows that: A. [Ag^+] = [SO_4^2-] B. [Ag^+] = (K_sp/4)^1/3 C. [Ag^+] = 2[SO4^2-] D. [Ag^+] = 1/2[SO4^2-] E. [Ag^+] = K_sp

Explanation / Answer

1) [OH-] = 0.015 , assuming Ca(NO3)2 completely dissociated we get [Ca2+] = 0.001 M

Now Ca2+] [OH-]^2 = ( 0.001) ( 0.015^2) = 2.25 x 10^ -7

Ionic product < Ksp , hence solution is unsaturated no preicpitate forms

2) Ag2SO4(s) <--> 2Ag+(aq) + SO42- (aq)

for S moles dissociation we get 2S of Ag+ and S of SO42-

at saturated solution [Ag+] = 2 x [SO42-]

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