A 1.00 *10^-3 mol L-1 solution of NaCl (aq) was added slowly to 50.0 mL of a wel
ID: 514785 • Letter: A
Question
A 1.00 *10^-3 mol L-1 solution of NaCl (aq) was added slowly to 50.0 mL of a well stirred 1.00*10^-3 mol L-1 solution of AgNO3 (aq). The following reaction occurred:Ag+ (aq) + Cl- (aq) ----> AgCl (s)
After addition of what volume of the NaCl (aq) solution will AgCl (s) first be observed in the reaction mixture? Ksp (AgCl) = 1.80 *10^-10
The answer that you should get is 9.00*10^-3 mL. Please show working, thanks! A 1.00 *10^-3 mol L-1 solution of NaCl (aq) was added slowly to 50.0 mL of a well stirred 1.00*10^-3 mol L-1 solution of AgNO3 (aq). The following reaction occurred:
Ag+ (aq) + Cl- (aq) ----> AgCl (s)
After addition of what volume of the NaCl (aq) solution will AgCl (s) first be observed in the reaction mixture? Ksp (AgCl) = 1.80 *10^-10
The answer that you should get is 9.00*10^-3 mL. Please show working, thanks!
Ag+ (aq) + Cl- (aq) ----> AgCl (s)
After addition of what volume of the NaCl (aq) solution will AgCl (s) first be observed in the reaction mixture? Ksp (AgCl) = 1.80 *10^-10
The answer that you should get is 9.00*10^-3 mL. Please show working, thanks!
Explanation / Answer
Ksp of AgCl = [Ag+][Cl-]
with [Ag+] = 1.00 x 10^-3 M
[Cl-] needed = 1.80 x 10^-10/1 x 10^-3
= 1.8 x 10^-7 M
So, it will react with equal amount of Ag+
Cl- moles needed = molarity x volume
= 1.8 x 10^-7 M x 50 ml
= 9.0 x 10^-6 mmol
therefore Volume of NaCl needed for precipitation of AgCl = moles/molarity
= 9.0 x 10^-6 mmol/1 x 10^-3 M
= 9.0 x 10^-3 ml
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