When 1.0 mL of 0.10 mol/L AgNO_3 was added to 4.0 mL of 2.0 mol/L Na_2 CO_3, the
ID: 507683 • Letter: W
Question
When 1.0 mL of 0.10 mol/L AgNO_3 was added to 4.0 mL of 2.0 mol/L Na_2 CO_3, the product formed in the reaction was Ag_2CO_3(s). From the measured potential of the cell, the calculated concentration of the silver ion in solution was determined to be less than 10^-7 mol/L. Calculate the concentration of the CO^2-_3 remaining in the solution after reaction? Answer: 1.6 mol/L In order to study the complex ion equilibrium: M^3+(aq) 2 Cl^-(aq) equilibrium [MCl_2]^+(aq), 5.0 mL, of 0.010 mol/L M^3+ was added to 5.0 mL of 1.0 mol/L NaCl. From the measured cell potential, the calculated [M^3+] was determined to be less than 10^-7 mol/L. What is the concentration of [MCl_2]^+(aq) at equilibrium? Answer: 5.0 middot 10^-3 mol/L.Explanation / Answer
Solution:- (1)
mmol of AgNO3 = 1.0 x 0.10 = 0.10 mmol
mmol of Na2CO3 = 4.0 x 2.0 = 8.0 mmol
Ag+ will come from AgNO3 and it will be 0.10 mmol
CO32- will come from Na2CO3 and it will be 8.0 mmol
From cell potential measured cell concentration of Ag+ is less than 10-7 mol/L which too low and indicates mostly all the Ag+ was used up.
Total volume = 1.0 ml + 4.0 ml = 5.0 ml
So, initial Ag+ = 0.10 mmol/5.0ml = 0.02 M
initial CO32- = 8.0mmol/5.0 ml = 1.6 M
2Ag+(aq) + CO32-(aq) -----> Ag2CO3(s)
I 0.02 1.6 0
C -0.02 0.01 0
E 0 (1.6 - 0.01) 0
From the ice table CO32- = 1.6 - 0.01 = 1.59 that could be round to 1.6 M which is same as the given answer.
(2) It could be solve similar to the previous one.
mmol of M3+ = 5.0 x 0.010 = 0.050 mmol
mmol of Cl- = 5.0 ml x 1.0 = 5.0 mmol
Total volume = 10.0 ml
Initial M3+ = 0.050mmol/10.0 ml = 0.0050 M
initial Cl- = 5.0 mmol/10.0 ml = 0.50 mmol
M3+(aq) + 2Cl-(aq) <-----> [MCl2]+(aq)
I 0.0050 0.50 0
C -0.0050 -0.01 +0.0050
E 0 0.49 0.0050
From the ice table M3+ is looks 0 but in actual it is greater than zero but too low even lower than 10-7 M
[MCl2]+ is 0.0050 that is 5.0 x 10-3 M.
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