Question

A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 24.7 mLmL of 1.50 MM H2SO4H2SO4 was needed

Answer 1

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.