Q1 . A solution of 0.3040 M KOH is used to neutralize 20.00mL of a H3PO4 solutio

Question

Q1. A solution of 0.3040 M KOH is used to neutralize 20.00mL of a H3PO4 solution.

H3PO4(aq) + 3KOH(aq) K3PO4(aq) + 3H2O(l)

If 32.52 mL of the KOH solution is required to reach the endpoint, what is the molarity of the H3PO4 solution?

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Q2. Sodium oxide, Na2O, reacts with water to give NaOH.

Part A

Write a balanced equation for the reaction.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Part B

What is the pH of the solution prepared by allowing 1.55 g of Na2O to react with 700.0 mL of water? Assume that there is no volume change.

Express your answer using three decimal places.

Part C

How many milliliters of 0.0100 M HCl are needed to neutralize the NaOH solution prepared in the part B?

Express your answer using three significant figures.

Explanation / Answer

Q1)

H3PO4(aq) + 3KOH(aq) K3PO4(aq) + 3H2O(l)

no.of moles of KOH = molarity*vol. im lt = 0.3252*0.3040 = 0.09886 moles

as per the balanced equation
3 moles of KOH rtequires ........... 1mole of H3PO4
0.09886 moles of KOH ....................?
      = 0.09886/3 = 0.032954 moles of H3PO4

Conc. of H3PO4 = 0.032954*0.020 = 6.6*10^-3 M

Q2)
PartA

Na2O(s) + H2O(l) ........>2NaOH (aq)

PartB

M = (wt/mol.wt)*(1000/vol. in ml)
M = (1.55/62)*(1000/700)
M = 0.036 M

[OH-] = 0.036 M
ph = 14 - Poh
   = 14 - [ -log(OH-)]
    = 14 - [-log(0.036)]
    = 12.55
PartC
M1V1 = M2V2
0.01*V1 = 0.036*700
V1 =2.5210^3 ml

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