What is the concentration of potassium ions in a solution obtained by mixing 480. mL of 0.410 M potassium hydroxide(aq) with 440. mL of 0.290 M potassium sulfate(aq). Enter your answer in decimal notation rounded to the appropriate number of significant figures.

The concentration of potassium ions, K⁺ in the solution obtained by mixing 480 mL of 0.410 M KOH with 440 mL of 0.290 M K₂SO₄ is 0.491 M

We'll begin by calculating the number of mole of potassium ion, K⁺ in each solution.

For KOH:

Volume = 480 mL = 480 / 1000 = 0.48 L

Molarity = 0.410 M

Mole of KOH =?

Mole = Molarity x Volume

Mole of KOH = 0.410 × 0.48

Mole of KOH = 0.1968 mole

KOH(aq) —> K⁺(aq) + OH¯(aq)

From the balanced equation above,

1 mole of KOH contains 1 mole of K⁺

Therefore,

0.1968 mole of KOH will also contain 0.1968 mole of K⁺

Thus, 0.1968 mole of K⁺ is present in 480 mL of 0.410 M KOH

For K₂SO₄:

Volume = 440 mL = 440 / 1000 = 0.44 L

Molarity = 0.290 M

Mole of K₂SO₄ =?

Mole = Molarity x Volume

Mole of K₂SO₄ = 0.290 × 0.44

Mole of K₂SO₄ = 0.1276 mole

K₂SO₄(aq) —> 2K⁺(aq) + SO₄²¯(aq)

From the balanced equation above,

1 mole of K₂SO₄ contains 2 moles of K⁺

Therefore,

0.1276 mole of K₂SO₄ will contain = 0.1276 × 2 = 0.2552 mole of K⁺

Thus, 0.2552 mole of K⁺ is present in 440 mL of 0.290 M K₂SO₄

Next, we shall determine the total mole of K⁺ in the resulting solution.

Mole of K⁺ in KOH = 0.1968 mole

Mole of K⁺ in K₂SO₄ = 0.2552 mole

Total mole = 0.1968 + 0.2552

Total mole = 0.452 mole

Next, we shall determine the total volume of the resulting solution

Volume of KOH = 0.48 L

Volume of K₂SO₄ = 0.44 L

Total volume = 0.48 + 0.44

Total volume = 0.92 L

Finally, we shall determine the molarity of K⁺ in the resulting solution

Total mole = 0.452 mole

Total volume = 0.92 L

Molarity of K⁺ =?

Molarity = mole / Volume

Molarity of K⁺ = 0.452 / 0.92

Molarity of K⁺ = 0.491 M

Therefore, the molarity of potassium ion, K⁺ in the resulting solution is 0.491 M

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