What Is the Original Concentration of NaOH in the Titration Lab?

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SUMMARY

The original concentration of NaOH in the titration lab can be calculated by analyzing the neutralization reactions involving H2SO4, NaOH, and KOH. The reaction shows that 0.01051 moles of H2SO4 react with an unknown amount of NaOH and 0.0009372 moles of KOH. To find the original molarity of NaOH, one must account for the total moles of base required to neutralize the sulfuric acid, which involves determining the moles contributed by NaOH and KOH.

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Homework Statement


A sol;n of 14.05 mL of NaOH is mixed with 50.0 mL of 0.2102 M H2SO4 (standardized) and allowed to react. The sol'n (H2SO4 and NaOH) is still acidic, so it is titrated with 10.55mL of standardized KOH (0.08883M) until the equilivance point is reached. What was the original concentration (molarity) of NaOH.


Homework Equations


I have converted H2SO4 to moles.
.2102 mol/L * .0500L = 0.01051 moles H2SO4

0.08883 mol/L KOH * 0.01055L = 0.0009372 mol KOH


The Attempt at a Solution


I am not sure how to link the 3 together
 
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Show us the reaction you are using for the neutralizations.

Basically (not jokingly), you are using two different bases to neutralize the sulfuric acid. Some moles of base are from KOH, and some unknown but calculable moles are from NaOH. HOW MANY BASE MOLES ARE NEEDED FOR THE NEUTRALIZATION? THis becomes just accounting.
 

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