# 0.7026g sample of unknown acid requires 40.96mL of 0.1158M NaOH for neutralization to a phenolphlein end point

There are 1.22mL of 0.2036M HCl used for back- titration.
How many moles of OH- are used? How many moles of H+ from HCl?
How many moles of H+ are there in the solid acid?
What is the molar mass of the unknown acid?

Moles HCl = 1.22 x 0.2036 /1000 = 0.000248
Moles NaOH = 40.96 x 0.1158 /1000 = 0.00474
Moles NaOH needed for titration = 0.00474 - 0.000248 = 0.00449
In the solid acid there are 0.00449 moles of H+
Molar mass = 0.7026 g / 0.00449 = 156.4 g

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