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?
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) + PO43-(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]
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