Question

Suppose 2.90g of nickel(II) iodide is dissolved in 150ml of a 0.70M aqueous solution of potassium carbonate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) iodide is dissolved in it. Round your answer to 3 significant digits.

Answer 1

Answer:

0.124 M.

Explanation:

Hello!

In this case, since the nickel iodide has the following formula:

NiI₂

So its molar mass is 312.5023 g/mol, in order to compute the molarity of the iodide anion, we first need the moles in 2.90 g:

$n_{NiI_2}=2.90g*\frac{1molNiI_2}{312.5023 gNiI_2} =0.00927molNiI_2$

Now, since one mole of nickel(II) iodide contains two mole of iodide anions, we infer there are 0.0186 moles of iodide cations. Moreover, since the molarity is computed by dividing the moles of those ones by the volume of the solution in liters, 150 mL (0.150 L) as it does not change, it turns out:

$M=\frac{0.0186molI^-}{0.150L}\\\\M=0.124M$

Best regards!