a) Solid sodium hydroxide is slowly added to 150 mL of a 0.0610 M silver nitrate

Question

a) Solid sodium hydroxide is slowly added to 150 mL of a 0.0610 M silver nitrate solution. The concentration of hydroxide ion required to just initiate precipitation is  M.

b). Solid calcium acetate is slowly added to 125 mL of a 0.0530 M sodium sulfite solution. The concentration of calcium ion required to just initiate precipitation is M.

c). Solid calcium nitrate is slowly added to 75.0 mL of a 0.0532 M sodium phosphate solution. The concentration of calcium ion required to just initiate precipitation is  M.

Explanation / Answer

AgNO3= 150ml of 0.0160M

number moles of AgNo3= 0.0160Mx0.150L= 0.0024 mole

AgNO3 (aq)+ NaOH(s) ---------- AgOH(s) + NaNO3(aq)

AgOH------------- Ag+ + OH-

ksp = [Ag+][OH-]                       Ksp of AgOH=2.0x10^-8

2.0x10^-8 = 0.0024x[OH]

[OH]= 8.33x10^-6mole

[OH-] = 8.33x10^-6/0.150L=5.55x10^-5M

[OH-] = 5.55x10^-5 M

b)sodium sulfite = Na2So3 = 125ml of 0.0530M

number of mole of Na2SO3 = 0.0530x.125L=0.006625 moles

Na2So3(aq) + Ca(CH3COO)2(s) -------------- CaSo3(s) + 2 NaCH3COO(aq)

CaSO3 -------------- Ca+2 + So3-2

Ksp =[Ca+2][So3-2]                   Ksp of CaSO3= 1.3x10^-8

1.3x10^-8 = [Ca+2]x0.006625

[Ca+2] = 1.96x10^-6

[Ca+2]= 1.96x10^-6/0.125=1.568x10^-5M

[Ca+2]= 1.568x10^-5M

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