Question

A student pipettes 50.00 NL of 0.102 M of HCl(aq) into a sample of antacid.the student finds that it takes 11.29 mL of 0.0825 NaOH(aq) to reach the endpoint of the titration. How many moles of base from the antacid are in the sample?

Answer 1

I think that NL is mL (milliliters).

Find how many moles of HCl(aq) and the base there are:

$$n(\text{HCl})=0.102\cdot50\cdot10^{-3}=5.1\cdot10^{-3}\text{ mol}.\\ n(\text{NaOH})=0.0825\cdot11.29\cdot10^{-3}=9.3\cdot10^{-3}\text{ mol}.$$

The reaction:

$$\text{HCl}+\text{NaOH}=>\text{H}_2\text{O}+\text{NaCl}.$$

Since NaOH reacts with HCl on a 1:1 mole basis, there were

$$n=9.3-5.1=4.2\text{ mmol}$$

of acid in the sample.