Question

Calculate the molarity of scandium nitrate and of Sc₃⁺(aq) ions and NO₃⁻ (aq) ions in a solution containing 2.86 grams of scandium nitrate dissolved in 250.0 milliliters of solution.

Answer 1

The molarity of scandium nitrate (Sc(NO3)3) in the solution is 0.0496 M. The molarity of scandium ions (Sc3+) is the same, 0.0496 M, while the molarity of nitrate ions (NO3-) is three times higher, 0.1488 M.

Step-by-step explanation:

To calculate the molarity of scandium nitrate in a solution, we first need to determine the number of moles of scandium nitrate present. Molarity is defined as moles of solute per liter of solution. Using the molecular weight of scandium nitrate (Sc(NO3)3), which is 230.948 g/mol, we can find out the number of moles:

Number of moles = mass (g) / molecular weight (g/mol) = 2.86 g / 230.948 g/mol ≈ 0.0124 moles of scandium nitrate.

Now, since the solution volume is 250.0 mL, which is 0.250 liters, the molarity of scandium nitrate is:

Molarity of Sc(NO3)3 = moles / volume (L) = 0.0124 moles / 0.250 L = 0.0496 M.

For Sc3+ ions, the molarity will be the same as that of Sc(NO3)3 since one formula unit of scandium nitrate yields one scandium ion. Therefore, [Sc3+] = 0.0496 M.

For NO3− ions, since one formula unit of scandium nitrate produces three nitrate ions, the molarity will be three times that of Sc(NO3)3:

[NO3−] = 3 × [Sc3+] = 3 × 0.0496 M = 0.1488 M.