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:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.01M\\V_2=25mL[/tex]
Putting values in above equation, we get:
[tex]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[/tex]
Hence, the volume of HCl neutralized is 0.125 mL
