Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na2CrO4 are added to 250 mL of 0.12 M AgNO3? [Ksp(Ag2CrO4) = 1.1 „e 10¡V12] What is the concentration of the silver ion remaining in solution?

To determine if a precipitate will form when 250 mL of 0.33 M Na2CrO4 are added to 250 mL of 0.12 M AgNO3, we can follow these steps:

1. Write the balanced chemical equation for the reaction between Na2CrO4 and AgNO3:
Na2CrO4 + 2AgNO3 -> Ag2CrO4 + 2NaNO3

2. Calculate the moles of Na2CrO4 and AgNO3:
Moles of Na2CrO4 = volume (in L) x concentration (in M)
= 0.250 L x 0.33 M
Moles of AgNO3 = volume (in L) x concentration (in M)
= 0.250 L x 0.12 M

3. Determine the limiting reagent by comparing the moles of Na2CrO4 and AgNO3. The reactant with fewer moles is the limiting reagent.
From step 2, we have:
Moles of Na2CrO4 = 0.250 L x 0.33 M = 0.0825 mol
Moles of AgNO3 = 0.250 L x 0.12 M = 0.030 mol

Na2CrO4 is the limiting reagent because it has fewer moles.

4. Use the stoichiometry of the balanced equation to determine the moles of Ag2CrO4 formed.
From the balanced equation, we see that 1 mole of Na2CrO4 reacts to form 1 mole of Ag2CrO4.
Therefore, the moles of Ag2CrO4 formed will also be 0.0825 mol.

5. Calculate the concentration of Ag+ ions remaining in solution:
The Ag2CrO4 produced will dissociate into 2Ag+ ions and 1CrO4^2- ion.
From the balanced equation, we see that 1 mole of Ag2CrO4 produces 2 moles of Ag+.
Therefore, the moles of Ag+ ions produced will be 2 x 0.0825 mol = 0.165 mol.

To find the concentration of Ag+ ions remaining in solution, we need to calculate the total volume of the solution:
Total volume = volume of Na2CrO4 + volume of AgNO3
= 0.250 L + 0.250 L
= 0.500 L

Concentration of Ag+ ions remaining = moles of Ag+ ions / total volume
= 0.165 mol / 0.500 L
= 0.33 M

Therefore, the concentration of the silver ion remaining in solution is 0.33 M.

To determine if a precipitate will form when 250 mL of 0.33 M Na2CrO4 is added to 250 mL of 0.12 M AgNO3, we need to compare the values of the ion product (IP) with the solubility product constant (Ksp).

The ion product (IP) is calculated by multiplying the concentrations of the ions present in the solution. In this case, the ions are Ag+ and CrO4^2-.

For AgNO3, the concentration of Ag+ is given as 0.12 M.

Na2CrO4 dissociates into three ions: 2Na+ and 1CrO4^2-. The concentration of CrO4^2- can be calculated as follows:

Since the initial volume of both solutions is equal, when mixed, we have a total volume of 500 mL (250 mL + 250 mL) or 0.5 L.

Concentration of CrO4^2- = (moles of Na2CrO4) / (total volume)

First, calculate the moles of Na2CrO4:

moles of Na2CrO4 = concentration of Na2CrO4 * volume of Na2CrO4 solution

moles of Na2CrO4 = 0.33 M * (250 mL / 1000 mL/L) = 0.0825 moles

Now, calculate the concentration of CrO4^2-:

Concentration of CrO4^2- = (0.0825 moles) / (0.5 L) = 0.165 M

So, the concentration of CrO4^2- is 0.165 M.

Now, we can calculate the ion product (IP) by multiplying the concentrations of Ag+ and CrO4^2-:

IP = [Ag+] * [CrO4^2-] = (0.12 M) * (0.165 M) = 0.0198

The solubility product constant (Ksp) of Ag2CrO4 is given as 1.1 x 10^-12.

Comparing the IP and Ksp values:

IP (0.0198) > Ksp (1.1 x 10^-12)

Since the IP is greater than Ksp, a precipitate will form. Both answers are correct.

To find the concentration of the remaining silver ions in solution, we need to determine how much Ag2CrO4 has precipitated. Let's assume x mol of Ag2CrO4 precipitates.

Since Ag2CrO4 dissociates into two Ag+ ions, the concentration of Ag+ that remains in solution can be calculated using the following equation:

[Ag+] = [initial Ag+] - [Ag+ that precipitates]

[Ag+] = 0.12 M - (2 * x)

Note: We multiply x by 2 because Ag2CrO4 dissociates into two Ag+ ions.

We can determine the value of x by using the stoichiometry of the balanced equation or by comparing the initial concentrations of Ag+ and CrO4^2-. However, this information is not provided in the question.

Therefore, we cannot determine the exact concentration of the silver ion remaining in solution without additional information.

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