Question

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III) nitrate solution. What is the final concentration of chromium(III) cation ?

Answer 1

Answer: The molarity of
`Cr^(3+)` ions in the solution is 0.299 M

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}` .....(1)

• For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

`0.234=\frac{\text{Moles of chromium (III) acetate}* 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=(0.234* 26.2)/(1000)=0.00613mol`

1 mole of chromium (III) acetate
`(Cr(CH_3COO)_3)` produces 1 mole of chromium
`(Cr^(3+))` ions and 3 moles of acetate
`(CH_3COO^-)` ions

Moles of
`Cr^(3+)\text{ ions}=(1* 0.00613)=0.00613moles`

• For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

`0.461=\frac{\text{Moles of chromium (III) nitrate}* 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=(0.461* 10.7)/(1000)=0.00493mol`

1 mole of chromium (III) nitrate
`(Cr(NO_3)_3)` produces 1 mole of chromium
`(Cr^(3+))` ions and 3 moles of nitrate
`(NO_3^-)` ions

Moles of
`Cr^(3+)\text{ ions}=(1* 0.00493)=0.00493moles`

• For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

ions in the solution is 0.299 M