Question

You have two 466.0 mL aqueous solutions. Solution A is a solution of silver nitrate, and solution B is a solution of potassium chromate. The masses of the solutes in each of the solutions are the same. When the solutions are added together, a blood-red precipitate forms. After the reaction has gone to completion, you dry the solid and find that it has a mass of 331.8 g. (a) Calculate the concentration of the potassium ions in the original potassium chromate solution.(b) Calculate the concentration of the chromate ions in the final solution

Answer 1

The concentration of the potassium ions in the original potassium chromate solution is 4.2927 mol/L.

The concentration of the chromate ions in the final solution is 1.0731 mol/L.

Step-by-step explanation:

`K_2CrO_4+2AgNO_3\rightarrow Ag_2CrO_4+2KNO_3`

Volume of solution A i.e. solution of silver nitrate = 466.0 mL = 0.466 L

Volume of solution B i.e. solution of potassium chromate = 466.0 mL = 0.466 L

Moles of silver chromate =
`(331.8)/(331.73 g/mol)=1.0002 mol`

According to reaction , 1 mol of silver chromate is produce from 2 moles of silver nitrate.

Then, 1.0002 moles of silver chromate will be formed from:

`(1)/(2)* 1.0002 mol=0.5001 mol` of silver nitrate.

According to reaction , 1 mol of silver chromate is produce from 1 mole of potassium chromate.

Then, 1.0002 moles of silver chromate will be formed from:

`(1)/(1)* 1.0002 mol=1.0002 mol` of potassium chromate

`Concentration =(Moles)/(Volume (L))`

a) The concentration of the potassium ions in the original potassium chromate solution.

Volume of the original solution = 0.466 L

1 mol of potassium chromate dissociates into 2mol of potassium ions and 1 mol of chromate ions:

Moles of potassium ions = 2 × 1.0002 mol = 2.0004 mol

`[K^+]=(2.0004 mol)/(0.466 L)=4.2927 mol/L`

b) The concentration of the chromate ions in the final solution

Volume of the final solution = 0.466 L + 0.466 L

Moles of chromate ions = 1 × 1.0002 mol = 1.0002 mol

`[CrO_4^(2+)]=(1.0002 mol)/(0.466 L+0.466L)=1.0731 mol/L`