Answer: The concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.
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 [tex]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]
We are given:
Conversion factor: 1 L = 1000 mL
[tex]n_1=1\\M_1=?M\\V_1=0.105L=105mL\\n_2=2\\M_2=0.100M\\V_2=46.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 105=2\times 0.100\times 46.5\\\\M_1=0.088M[/tex]
Hence, the concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.
