Question

Suppose that 31.3 mL of 0.487 M NaCl is added to 31.3 mL of 0.245 M AgNO₃.

Write a balanced molecular equation for the reaction. Include physical states for all reagents and products.

How many moles of AgCl would precipitate?
What are the concentrations of each of the ions in the reaction mixture after the reaction occurs?

Answer 1

NaCl = Na+ + Cl-

AgNO3 = Ag+ + NO3-

note that Ag+ and Cl- can form precipitate

molecular

AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq)

Ag+(aq) + Cl-(aq) -> AgCl(s)

so

mol of NaCl = MV = 0.487*31.3 = 15.2431 mol

mol of AgNO3 = MV = 0.245*31.3 = 7.6685 mol

ratio is 1:1 so

mol of Ag+ = 7.6685, mol of Cl- = 14.24

there is excess Cl-

expect only 7.6685 mol of AgCl to form

After reaction:

Vtotal = 31.3+31.3 = 62.6

[Na+] = 0.487/2 = 0.2435 M

[NO3-] =  0.487/2 = 0.2435 M

[Cl-] = (15.2431 -7.6685 )/(31.3+31.3) = 0.121 M

[Ag+] = 0 (approx)

answered by: anonymous