Question

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)

Answer 1

Answer: The molarity of KOH is 0.84 M.

Explanation:

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.