Question

When 160. mL Fe(NO3)3 solution at 1.50 M and 90. mL Ca(NO3)2 solution at 2.40 M are mixed together, what is the concentration of NO3¯ of the mixture?

Answer 1

Answer: The concentration of nitrate ions in the mixture is 3.65 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$    .....(1)

• For iron (II) nitrate:

Molarity of iron (II) nitrate solution = 1.50 M

Volume of solution = 160. mL = 0.160 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

$1.50M=\frac{\text{Moles of iron (II) nitrate}}{0.160L}\\\\\text{Moles of iron (II) nitrate}=(1.50mol/L\times 0.160L)=0.24mol$

1 mole of iron (II) nitrate produces 2 moles of nitrate ions and 1 mole of iron (II) ions

Moles of nitrate ions $,n_1=(2\times 0.24)=0.48mol$

• For calcium nitrate:

Molarity of calcium nitrate solution = 2.40 M

Volume of solution = 90. mL = 0.090 L

Putting values in equation 1, we get:

$2.40M=\frac{\text{Moles of calcium nitrate}}{0.090L}\\\\\text{Moles of calcium nitrate}=(2.40mol/L\times 0.090L)=0.216mol$

1 mole of calcium nitrate produces 2 moles of nitrate ions and 1 mole of calcium ions

Moles of nitrate ions $,n_2=(2\times 0.216)=0.432mol$

Calculating the concentration of nitrate ions in the mixture by using equation 1:

Total moles of nitrate ions = [0.48 + 0.432] mol = 0.912 moles

Total volume of the solution = [0.160 + 0.090] L = 0.250 L

Putting values in equation 1, we get:

$\text{Molarity of nitrate ions in the mixture}=\frac{0.912mol}{0.250L}\\\\\text{Molarity of nitrate ions in the mixture}=3.65M$

Hence, the concentration of nitrate ions in the mixture is 3.65 M