# Cancerous Chem Question Help (1 Viewer)

#### Life'sHard

##### New Member
Ksp for AgCl = 1.76x10^-10

NaCl (0.25mol) is added with AgNO3 (1.00mol) in 1 L of water. Determine the concentration of the silver ion [Ag+]

• TheShy

#### TheShy

##### Member
Reaction: NaCl + AgNO3 --> AgCl + NaNO3

You know that NaCl is the limiting reagent, so use that to find the moles of AgCl, ie .25 mol. Therefore, moles of Cl^- in AgCl will also be 0.25 moles, therefore the conc of the Cl^-1 is 0.25 molL^-1.

Ksp=[Ag+][Cl-]
therefore, [Ag+]=ksp/[Cl-]
sub the values in, and you get conc of Ag+

• tejmehta3108

#### Life'sHard

##### New Member
Reaction: NaCl + AgNO3 --> AgCl + NaNO3

You know that NaCl is the limiting reagent, so use that to find the moles of AgCl, ie .25 mol. Therefore, moles of Cl^- in AgCl will also be 0.25 moles, therefore the conc of the Cl^-1 is 0.25 molL^-1.

Ksp=[Ag+][Cl-]
therefore, [Ag+]=ksp/[Cl-]
sub the values in, and you get conc of Ag+
So would [Ag+]= Ksp/0.25

• TheShy

#### CM_Tutor

##### Moderator
Moderator
We start with [Ag+] = 1.00 mol L-1 and [Cl-] = 0.250 mol L-1

The Ksp equilibrium is

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

and it lies far to the left given the small value of Ksp = [Ag+][Cl-] = 1.76 x 10-10

So, at equilibrium, we will have [Ag+] = 1.00 - x mol L-1 and [Cl-] = 0.250 - x mol L-1

and for the equation (1.00 - x)(0.250 - x) = 1.76 x 10-10 to be solvable, noting x is the extent to which the system moved left (and hence x < 0.250), we will have [Cl-] = being very close to 0, x very close to 0.250, and so [Ag+] very close to 0.750 mol L-1.

At equilibrium, [Ag+] = 0.750 mol L-1 (3 sig. fig.)

[Cl-] = Ksp / [Ag+] = 1.76 x 10-10 / 0.750 = 2.34 x 10-10 mol L-1 (3 sig. fig.)

Check:
x = 0.250 - [[Cl-] = 0.249 999 999 76... mol L-1

and so actually [Ag+] = 0.750 000 000 23... mol L-1, which is consistent with the approximation above that [Ag+] = 0.750 mol L-1 (3 sig. fig.)

which fits above results.

Edited to emphasise result for concentration of silver(I) ions.

Last edited:

#### Life'sHard

##### New Member
We start with [Ag+] = 1.00 mol L-1 and [Cl-] = 0.250 mol L-1

The Ksp equilibrium is

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

and it lies far to the left given the small value of Ksp = [Ag+][Cl-] = 1.76 x 10-10

So, at equilibrium, we will have [Ag+] = 1.00 - x mol L-1 and [Cl-] = 0.250 - x mol L-1

and for the equation (1.00 - x)(0.250 - x) = 1.76 x 10-10 to be solvable, noting x is the extent to which the system moved left (and hence x < 0.250), we will have [Cl-] = being very close to 0, x very close to 0.250, and so [Ag+] very close to 0.750 mol L-1.

At equilibrium, [Ag+] = 0.750 mol L-1 (3 sig. fig.)

[Cl-] = Ksp / [Ag+] = 1.76 x 10-10 / 0.750 = 2.34 x 10-10 mol L-1 (3 sig. fig.)

Check:
x = 0.250 - [[Cl-] = 0.249 999 999 76... mol L-1

and so actually [Ag+] = 0.750 000 000 23... mol L-1

which fits above results.
But what's the concentration for Ag+ ?
Gotta dumb this down for me lol.

#### CM_Tutor

##### Moderator
Moderator
@Life'sHard, I have added bolded and coloured red emphasis to my original post.

• Life'sHard

#### Life'sHard

##### New Member
@Life'sHard, I have added bolded and coloured red emphasis to my original post.
So "TheShy" answer is incorrect I presume?

#### CM_Tutor

##### Moderator
Moderator
So "TheShy" answer is incorrect I presume?
The 0.25 mol of chloride ions must be split between the solid AgCl and the small amount in solution - they can't both be 0.25 mol.

As my answer shows, the [Cl-] in the solution is around 2.34 x 10-10 mol in the 1 L of solution, and the n(AgCl) precipitated as solid is 0.249 999 999 76... mol, which add to 0.25 mol

• Life'sHard

#### Life'sHard

##### New Member
The 0.25 mol of chloride ions must be split between the solid AgCl and the small amount in solution - they can't both be 0.25 mol.

As my answer shows, the [Cl-] in the solution is around 2.34 x 10-10 mol in the 1 L of solution, and the n(AgCl) precipitated as solid is 0.249 999 999 76... mol, which add to 0.25 mol
Ahhh I see tyvm