Answer: The volume of HCl required will be 28.57 mL.
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 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:
\(n_1=1\\M_1=3.5M\\V_1=?mL\
_2=1\\M_2=2M\\V_2=50mL\)
Putting values in above equation, we get:
\(1\times 3.5\times V_1=1\times 2\times 50\\\\V_1=28.57mL\)
Hence, the volume of HCl required will be 28.57 mL.
