Calculate how many moles of Na2S2O3 must be added to dissolve 0.020 mol AgBr in 1.0 L of water at 25°C (assume constant volume).

AgBr(s) ↔ Ag+(aq) + Br−(aq)
Ksp = 3.3 x 10^−13

Ag+(aq) + 2S2O32−(aq) ↔ [Ag(S2O3)2]3−(aq)
Kf = 2.0 x 10^13

(1) 0.0078 mol (2) 0.048 mol (3) 0.040 mol (4) 0.024 mol (5) 0.016 mol

Your Ksp for AgBr is eqn 1.

Your Kf for Ag(S2O3^-2)^3- is eqn 2.

Add equn 1 to eqn 2 to get
AgBr(s) + 2(S2O3)^3- ==> Br^- + [Ag(S2O3)2]^3-
Write the equilibrium expressiion for that which equals Ksp*Kf.
Then substitute 0.02 for Br^- and 0.02 for [Ag(S2O3)]^3- and solve for (S2O3)^2-. Remember AgBr(s) does not appear in the equilibrium expression. Looks like (1) is the answer but you should confirm that yourself.

Well, this is a chemistry question. Let me mix some humor into this chemical equation!

First, let's laugh at the AgBr(s), it's like Ag going "Hey Br, I'm gonna be positive today, are you down with that?" And Br just casually says "Positive or negative, it makes no difference to me. I'm all about that chemical flexibility, baby!"

Now, to dissolve this 0.020 mol AgBr in 1.0 L of water, we need the help of Na2S2O3. Let's see how many moles it takes to do the job:

The equation tells us that 1 mole of AgBr reacts with 1 mole of Ag+ and Br-. But wait, we don't have AgBr+ yet, we need S2O32- to make that happen.

According to the equation, 1 mole of Ag+ reacts with 2 moles of S2O32-. So, to dissolve 0.020 mol of AgBr, we would need twice that amount of Na2S2O3, which is 0.040 mol.

So, the answer is option (3) 0.040 mol. Na2S2O3 better be ready to lend a helping mole to the AgBr! Keep those chemicals laughing!

To solve this problem, we need to use the equilibrium constant expression for the formation of [Ag(S2O3)2]3−(aq):

[Ag+(aq)] * [S2O32−(aq)]^2 / [Ag(S2O3)2]3−(aq)

Let's start by calculating the concentration of [Ag+] in the solution.

Given:
Moles of AgBr = 0.020 mol
Volume of solution = 1.0 L

Since AgBr dissociates into Ag+ and Br− in a 1:1 ratio, the concentration of Ag+ is equal to the moles of AgBr divided by the volume of the solution.

[Ag+] = 0.020 mol / 1.0 L
[Ag+] = 0.020 M

Now, let's calculate the concentration of [S2O32−] using the Kf value and the concentration of [Ag+].

Kf = [Ag(S2O3)2]3−(aq) / ([Ag+] * [S2O32−]^2)

Since we're assuming constant volume, the concentration of [Ag+] is the same as the concentration of AgBr added.

0.020 M = [Ag(S2O3)2]3−(aq) / (0.020 M * [S2O32−]^2)

Simplifying, we get:

[S2O32−]^2 = [Ag(S2O3)2]3−(aq) / (0.020 M * 0.020 M)

Now, we need to calculate the concentration of [Ag(S2O3)2]3−(aq) using the Ksp value and the concentration of [Ag+].

Ksp = [Ag+(aq)] * [Br−(aq)]

Given that Ksp = 3.3 x 10^−13 and [Ag+] = 0.020 M, we can solve for [Br−(aq)]:

3.3 x 10^−13 = 0.020 M * [Br−(aq)]

[Br−(aq)] = (3.3 x 10^−13) / 0.020 M
[Br−(aq)] = 1.65 x 10^−11 M

Now that we have the concentration of [Br−(aq)], we can substitute it back into the equation for [S2O32−]^2:

[S2O32−]^2 = [Ag(S2O3)2]3−(aq) / (0.020 M * (1.65 x 10^−11 M))

Finally, to find the moles of Na2S2O3 needed, we can use the balanced equation:

2Na2S2O3(aq) + 3Br−(aq) → [Na3B(S2O3)3](aq) + 6Na+(aq)

According to the equation, 2 moles of Na2S2O3 react with 3 moles of Br−(aq) to form the complex [Na3B(S2O3)3](aq). Therefore, the moles of Na2S2O3 required is:

Moles of Na2S2O3 = (1.0 L) * [S2O32−]^2
Moles of Na2S2O3 = (1.0 L) * ([Ag(S2O3)2]3−(aq) / (0.020 M * (1.65 x 10^−11 M))) * 2

Now, we need the value of [Ag(S2O3)2]3−(aq) to substitute in the equation:

Kf = [Ag(S2O3)2]3−(aq) / (0.020 M * [S2O32−]^2)

Since we don't have the exact value of [Ag(S2O3)2]3−(aq), we cannot calculate the exact value for the moles of Na2S2O3.

To find the number of moles of Na2S2O3 needed to dissolve 0.020 mol of AgBr, we need to consider the equilibrium reactions involved.

First, let's balance the equation for the dissolution of AgBr:

AgBr(s) ↔ Ag+(aq) + Br−(aq)

Next, let's calculate the concentration of Ag+ ions in the solution at equilibrium. Since we know that the volume of the solution is 1.0 L, the concentration of Ag+ ions is equal to the number of moles of AgBr divided by the volume:

[Ag+(aq)] = 0.020 mol / 1.0 L = 0.020 M

Now, let's determine the concentration of S2O32− ions needed to form a complex with Ag+ ions using the formation constant Kf.

[Ag+(aq)] + 2[S2O32−(aq)] ↔ [Ag(S2O3)2]3−(aq)

The concentration of S2O32− ions required is equal to half of the concentration of Ag+ ions, because the stoichiometric coefficient of S2O32− in the complex is 2. Therefore:

[S2O32−(aq)] = 0.5 * [Ag+(aq)] = 0.5 * 0.020 M = 0.010 M

Finally, let's calculate the number of moles of Na2S2O3 needed to achieve this concentration of S2O32− ions. Since the molar ratio between Na2S2O3 and S2O32− is 1:2, the number of moles of Na2S2O3 is:

moles of Na2S2O3 = [S2O32−(aq)] * volume = 0.010 M * 1.0 L = 0.010 mol

Therefore, the correct answer is (3) 0.040 mol.