Question

If 100. mL of 0.100 M K2SO4 is added to 200. mL of 0.150 M KCl, what is the concentration of K + ions in the final solution? Assume that the volumes are additive.

Answer 1

Concentration = (2 * 0.1 * 100 + 1 * 0.150 * 200) / (200+100)

Concentration = 0.167 M of K+ ions

Answer 2

The final concentration will be "0.167 M".

According to the question,

Moles of KCl,

• $0.15\times 200\times 10^{-3}$

Moles of K₂SO₄,

• $0.1\times 100\times 10^{-3}$

Total moles of K+ will be:

= $0.15\times 200\times 10^{-3}+2\times 0.1\times 100\times 10^{-3}$

Total volume will be:

= $(100+200) \ mL$

hence,

→ The final concentration of K+ will be:

= $\frac{Total \ moles}{Total \ volume}$

= $\frac{0.05}{300}\times 100$

= $\frac{0.5}{3}$

= $0.167 \ M$

Thus the above solution is correct.

Learn more:

https://brainly.com/question/7466212