Question

Please show all work.

1.) What is the concentration of potassium ions in 100. mL of a 0.21M solution of K3PO4?

2.) What volume of 0.150M iron (11) sulfate is required to produce a 250mL solution that is 0.045M iron (11) sulfate?

3.) What volume of 0.70M KOH is needed to neutralize 25.0mL of 0.55M HCl?

Answer 1

1)

[K₃PO₄] = 0.21 M

The volume of K₃PO₄ solution = 100. mL x ( 1L / 1000 mL) = 0.1 L

Determine the number of moles of K₃PO₄ using the concentration and volume as follows:

The relation between molarity, Volume of the solution, and the number of moles is as follows:

Molarity = Number of moles / Volume of solution (L)

Rearrange the formula for the number of moles as follows:

Number of moles = Molarity x Volume of solution(L)

Substitute 0.21 M for the molarity and 0.1 L for the volume of the solution. Determine the Number of moles as follows:

Number of moles = 0.21 M x 0.1 L

Number of moles = 0.021 mol K₃PO₄

The dissociation of K₃PO₄ is as follows:

K₃PO₄(aq) 3K⁺(aq) + PO₄^3-(aq)

Use the moles of K₃PO₄ and the mole ratio from the above dissociation equation. Determine the number of moles of potassium ions as follows:

= 0.021 mol K₃PO₄ x ( 3 mol K⁺ / 1 mol K₃PO₄)

= 0.063 mol K⁺

Determine the concentration of potassium ions in 100 mL solution as follows:

Molarity = Number of moles / Volume of solution (L)

Substitute 0.063 mol for the number of moles and 0.1 L for the volume of the solution. Determine the concentration of potassium ions as follows:

Molarity = 0.063 mol / 0.1 L

Molarity = 0.63 M

Thus, the concentration of potassium ions in 100 mL of 0.21 M solution of K₃PO₄ is 0.63 M.

2)

Concentration of iron(II) sulfate solution(M₁) = 0.045 M

Volume of iron(II) sulfate solution(V₁) = 250 mL

Concentration of iron(II) sulfate solution(M₂) = 0.150 M

The dilution formula is as follows:

M₁V₁ = M₂V₂

Rearrange the formula for V₂ as follows:

V₂ = (M₁V₁) / M₂

Substitute 0.045 M for M₁, 250 mL for V₁, 0.150 M for M₂. Determine the volume of stock solution required as follows:

V₂ = (0.045 M x 250 mL ) / 0.150 M

V₂ = 75 mL

Thus, 75 mL of 0.150 M iron(II) sulfate solution is required to produce a 250 mL solution of 0.045 M iron(II) sulfate.

3)

Concentration of HCl(aq) = 0.55 M

Volume of HCl(aq) = 25.0 mLx (1 L / 1000 mL) = 0.025 L

Volume of KOH(aq) = 0.70 M

Determine the number of moles of HCl using the concentration and volume of the solution as follows:

The relation between molarity, Volume of the solution, and the number of moles is as follows:

Molarity = Number of moles / Volume of solution (L)

Rearrange the formula for the number of moles as follows:

Number of moles = Molarity x Volume of solution(L)

Substitute 0.55 M for the molarity and 0.025 L for the volume of the solution. Determine the Number of moles as follows:

Number of moles = 0.55 M x 0.025 L

Number of moles = 0.01375 mol HCl

The balanced neutralization reaction between HCl and KOH is as follows:

HCl(aq) + KOH(aq) KCl(aq) + H₂O(l)

Use the moles of HCl and the mole ratio from the balanced chemical reaction and determine the number of moles of KOH as follows:

= 0.01375 mol HCl x ( 1 mol KOH / 1 mol HCl)

= 0.01375 mol KOH

Use the concentration of KOH and moles of KOH. Determine the volume of the KOH solution as follows:

Molarity = Number of moles / Volume of solution (L)

Rearrange the formula for the volume of solution as follows:

The volume of solution(L) = Number of moles / Molarity

Substitute 0.01375 mol for the number of moles and 0.70 M for Molarity. Determine the volume of KOH solution as follows:

The volume of solution(L) = 0.01375 mol / 0.70 M

The volume of the solution(L) = 0.0196 L x ( 1000 mL / 1 L) = 19.6 mL

Thus, the volume of 0.70 M KOH needed to neutralize 25.0 mL of 0.55 M HCl is 20. mL [ 2 S.F]