Question

In the laboratory, a student adds 19.7 g of barium acetate to a 500. mL volumetric flask and adds water to the mark on the neck of the flask. Calculate the concentration (in mol/L) of barium acetate, the barium ion and the acetate ion in the solution. [Ba(CH3COO)2] = M [Ba2+] = M [CH3COO-] = M

Calculate the mass, in grams, of iron(II) sulfate that must be added to a 300-mL volumetric flask in order to prepare 300 mL of a 0.181 M aqueous solution of the salt.

grams

Answer 1

1)
`[Ba(CH_3COO)_2]=0.1545 mol/L`

`[Ba^(2+)]=0.1545 mol/L`

`[CH_3COO^-]=0.3090 mol/L`

2) 21.72 grams of iron(II) sulfate that must be added.

Step-by-step explanation:

`Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution (L)}}`

1) Moles of barium acetate =
`(19.7 g)/(255 g/mol)=0.07725 mol`

Volume of the solution was made to 500 ml that 0.5 L

`[Ba(CH_3COO)_2]=(0.07725 mol)/(0.5L)=0.1545 mol/L`

In 1 mole of barium acetate there are 1 mole of barium ions and 2 moles of acetate ions.

`[Ba^(2+)]=1* [Ba(CH_3COO)_2]`

`[Ba^(2+)]=1* 0.1545 mol/L=0.1545 mol/L`

`[CH_3COO^-]=2* [Ba(CH_3COO)_2]`

`[CH_3COO^-]=2* 0.1545 mol/l=0.3090 mol/L`

2) Moles of iron(II) sulfate be n

Volume of the solution = 300 mL= 0.3 L

`[Fe_2(SO_4)_3]=0.181 M`

`0.181 M=(n)/(0.3 L)`

n = 0.0543 moles

Mass of 0.0543 moles of iron(II) sulfate:

0.0543 mol × 400 g/mol = 21.72 g

21.72 grams of iron(II) sulfate that must be added.