Question

calculate the concentration of k+ if the following are mixed together in a beaker: 25.0 mL of 0.250 M KI, 25.0 mL of 0.100 K2SO4, and 15.0 mL of 0.100 M MgCl2

Answer 1

Molarity = Moles / Liter

Moles = Molarity x Liter

25.0 mL of 0.250 M KI

Moles of KI = 0.250 M x 0.025 L

= 0.00625 moles

KI K⁺ + I^-

1 mole of KI gives 1 mole of K⁺.

So, moles of K⁺ = 0.00625 moles

25.0 mL of 0.100 K2SO4

Moles of K2SO4 = 0.100 M x 0.025 L

= 0.0025 moles

K2SO4 2 K⁺ + SO₄^2-

1 mole of K2SO4 produces 2 moles of K⁺.

So, 0.0025 moles of K2SO4 produces 2 x 0.0025 moles of K⁺.

0.0025 moles of K2SO4 produces 0.0050 moles of K⁺.

15.0 mL of 0.100 M MgCl2 has no effect on moles of K⁺.

Total moles of K⁺ = 0.00625 moles + 0.0050 moles

= 0.01125 moles

Total volume = 25.0 mL + 25.0 mL + 15.0 mL

= 65 mL

= 0.065 L

Now,

[K⁺] = 0.01125 moles / 0.065 L = 0.173 M