Question

A volume of 40.0 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 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Answer 1

Answer: The concentration of KOH solution is 1.215 M

Explanation:

For the given chemical equation:

$2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)$

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:

$n_1=2\\M_1=1.50M\\V_1=16.2mL\\n_2=1\\M_2=?M\\V_2=40.0mL$

Putting values in above equation, we get:

$2\times 1.50\times 16.2=1\times M_2\times 40.0\\\\M_2=\frac{2\times 1.50\times 16.2}{1\times 40.00}=1.215M$

Hence, the concentration of KOH solution is 1.215 M