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

[engineer2010]

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

 

[symbolipoint]

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.