Exercise 7 If 60.2 mL of 0427 M KOH solution are required to neutralize 10.1 mL of H_(2)SO_(4) solution, what is the concentration of the H_(2)SO_(4) solution in molarity?

Question

Exercise 7 If 60.2 mL of 0427 M KOH solution are required to neutralize 10.1 mL of H_(2)SO_(4) solution, what is the concentration of the H_(2)SO_(4) solution in molarity?

Answer 1

To find the concentration of the

solution in molarity, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between KOH and

.The balanced chemical equation for the neutralization reaction is:

From the balanced equation, we can see that 2 moles of KOH react with 1 mole of

.Given information:- Volume of KOH solution: 60.2 mL- Concentration of KOH solution: 0.427 M- Volume of

solution: 10.1 mLStep 1: Calculate the number of moles of KOH.Number of moles of KOH = Concentration of KOH × Volume of KOH solutionNumber of moles of KOH = 0.427 M × 0.0602 L = 0.0257 molesStep 2: Calculate the number of moles of

.Since 2 moles of KOH react with 1 mole of

, the number of moles of

is half the number of moles of KOH.Number of moles of

= 0.0257 moles / 2 = 0.01285 molesStep 3: Calculate the concentration of the

solution.Concentration of

= Number of moles of

/ Volume of

solutionConcentration of

= 0.01285 moles / 0.0101 L = 1.27 MTherefore, the concentration of the

solution is 1.27 M.

Soru

Exercise 7 If 60.2 mL of 0427 M KOH solution are required to neutralize 10.1 mL of H_(2)SO_(4) solution, what is the concentration of the H_(2)SO_(4) solution in molarity?

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