Answer: Thus molarity of [tex]HNO_3[/tex] is 1.35 M
Explanation:
To calculate the volume 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 KOH.
We are given:
[tex]n_1=1\\M_1=?M\\V_1=10.0mL\\n_2=1\\M_2=1.0M\\V_2=13.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 10.0=1\times 1.0\times 13.5\\\\M_1=1.35M[/tex]
Thus molarity of [tex]HNO_3[/tex] is 1.35 M