Question

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 2 M was neutralized by 0.1 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 1.25 mL

Step-by-step explanation:

To calculate the volume 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 stomach acid which is HCl

`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=0.1M\\V_2=25mL`

Putting values in above equation, we get:

`1* 2* V_1=1* 0.1* 25\\\\V_1=(1* 0.1* 25)/(1* 2)=1.25mL`

Hence, the volume of HCl neutralized is 1.25 mL