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denis23 [38]
3 years ago
10

A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What

was the molarity of the KOH solution if 25.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
Chemistry
1 answer:
goblinko [34]3 years ago
7 0

<u>Answer:</u> The molarity of KOH is 0.84 M.

<u>Explanation:</u>

To calculate the concentration 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=1.50M\\V_1=25.2mL\\n_2=1\\M_2=?M\\V_2=90mL

Putting values in above equation, we get:

1\times 1.50\times 25.2=2\times M_2\times 90\\\\M_2=0.84M

Hence, the molarity of KOH is 0.84 M.

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