Question

g What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sample; this was the HCl remaining after the antacid tablet did it's job)

Answer 1

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

Answer: The volume of HCl neutralized is 0.125 mL

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl (Stomach acid)

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.01M\\V_2=25mL$

Putting values in above equation, we get:

$1\times 2\times V_1=1\times 0.01\times 25\\\\V_1=\frac{1\times 0.01\times 25}{1\times 2}=0.125mL$

Hence, the volume of HCl neutralized is 0.125 mL