What volume (L) of 0.236 M potassium hydroxide solution would just neutralize 49.3 mL of 0.217 M H2SO4 solution?

Chemistry by OpenStax (2015-05-04)
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ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter4: Stoichiometry Of Chemical Reactions
Section: Chapter Questions
Problem 78E: What volume of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100-M Ca(OH)2 solution?...
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What volume (L) of 0.236 M potassium hydroxide solution would just neutralize 49.3 mL of 0.217 M H2SO4 solution?

Expert Solution
Step 1

The reaction between H2SO4 and KOH is given as

H2SO4 + 2 KOH ---> K2SO4 + 2 H2O

Hence from the above balanced reaction we can say that the moles of KOH required = moles of H2SO4 to be neutralised X 2

Since moles of H2SO4 to be neutralised = molarity of H2SO4 solution X volume of H2SO4 solution in L = 0.217 X 49.3 / 1000

=>moles of H2SO4 to be neutralised = 0.0107 approx 

hence moles of KOH required = 0.0107 X 2 = 0.0214 

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