Answer :
The answer is (3) 0.193M. To find the molarity of HNO3, you just need to use the M1V1=M2V2 equation (no need to worry about ionization constants because HNO3 is monoprotic and KOH dissociates 1:1). Since the molarity you are looking for is M1, you get M1=M2V2/V1=(0.150)(32.1)/25.0= 0.193M
Answer: The correct answer is Option 3.
Explanation:
To calculate the molarity of the acid, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of acid
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of base
We are given:
[tex]M_1=?M\\V_1=25mL\\M_2=0.150M\\V_2=32.1mL[/tex]
Putting values in above equation, we get:
[tex]M_1\times 25=0.15\times 32.1\\\\M_1=0.193M[/tex]
Hence, the correct answer is Option 3.
