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

Answer:

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.

Explanation:

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

1) Moles of barium acetate = $\frac{19.7 g}{255 g/mol}=0.07725 mol$

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

$[Ba(CH_3COO)_2]=\frac{0.07725 mol}{0.5L}=0.1545 mol/L$

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

$[Ba^{2+}]=1\times [Ba(CH_3COO)_2]$

$[Ba^{2+}]=1\times 0.1545 mol/L=0.1545 mol/L$

$[CH_3COO^-]=2\times [Ba(CH_3COO)_2]$

$[CH_3COO^-]=2\times 0.1545 mol/l=0.3090 mol/L$

2) Moles of iron(II) sulfate be n

Volume of the solution = 300 mL= 0.3 L

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

$0.181 M=\frac{n}{0.3 L}$

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.