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GREYUIT [131]
3 years ago
9

A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2

SO4). What was the molarity of the KOHKOH solution if 24.7 mLmL of 1.50 MM H2SO4H2SO4 was needed
Chemistry
1 answer:
Alja [10]3 years ago
7 0

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O (Neutralization reaction)

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=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

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