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 ?

The molarity of ions in the solution is 0.299 M

Explanation:

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

   .....(1)

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

1 mole of chromium (III) acetate produces 1 mole of chromium ions and 3 moles of acetate ions

Moles of

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

1 mole of chromium (III) nitrate produces 1 mole of chromium ions and 3 moles of nitrate ions

Moles of

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