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Law Incorporation [45]
3 years ago
15

If 72.1 mL of 0.543 M H2SO4 completely titrates 39.0 mL of KOH solution, what is the molarity of the KOH solution? a. 0.317 M b.

0.502 M c. 1.00 M d. 2.01 M
Chemistry
1 answer:
Snezhnost [94]3 years ago
4 0

<u>Answer:</u> The correct answer is Option  d.

<u>Explanation:</u>

To calculate the molarity of base, 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 H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH

We are given:

n_1=2\\M_1=0.543M\\V_1=72.1mL\\n_2=1\\M_2=?M\\V_2=39.0mL

Putting values in above equation, we get:

2\times 0.543\times 72.1=1\times M_2\times 39.0\\\\M_2=2.01M

Hence, the correct answer is Option d.

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