RobtGodina935
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A 25.0-milliliter sample of HNO3(aq) is neutralized by 32.1 milliliters of 0.150 M KOH(aq). What is the molarity of the HNO3(aq)?
(1) 0.117 M (3) 0.193 M
(2) 0.150 M (4) 0.300 M



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.

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