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azamat
3 years ago
7

If 42.68 ml of 0.43 M KOH is used to neutralize 18.40 ml of H, SO4, calculate

Chemistry
1 answer:
Temka [501]3 years ago
8 0

The concentration of the sulfuric acid : 0.499 M

The net ionic equation

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

<h3>Further explanation</h3>

Given

42.68 ml of 0.43 M KOH

18.40 ml of H2SO4

Required

the concentration of the sulfuric acid

the net ionic equation

Solution

Acid-base titration formula  

Ma. Va. na = Mb. Vb. nb  

Ma, Mb = acid base concentration  

Va, Vb = acid base volume  

na, nb = acid base valence  

Input the value :

a=KOH, b= H2SO4

0.43 x 42.68 x 1 = Mb x 18.40 x 2

Mb = 0.499 M

The net ionic equation

Reaction

2KOH + H2SO4 → K2SO4 + 2H2O

2 K ⁺ (aq]+ 2 OH ⁻ (aq] + 2 H ⁺ (aq] + SO ₄²⁻(aq] → 2 K ⁺ (aq] + SO ₄²⁻(aq] + 2 H ₂ O (l]

canceled the spectator ions :

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

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Answer:

The answer to your question is 88.7 ml

Explanation:

Data

Volume = ?

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Balanced chemical reaction

               2 NaOH + H₂C₄H₄O₆  ⇒  Na₂C₄H₄O₆  +  2H₂O

1.- Calculate the moles of H₂C₄H₄O₆

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moles = Molarity x volume

Substitution

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Result

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2.- Use proportions to calculate the moles of NaOH

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               x = (0.0063 x 2) / 1

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3.- Calculate the volume  of NaOH

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Solve for volume

Volume = moles/Molarity

Substitution

Volume = 0.0126/0.142

Result

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