Question

In an acid-base neutralization reaction 43.74 ml of 0.500 m potassium hydroxide reacts with 50.00 ml of sulfuric acid solution. what is the concentration of the h2so4 solution?'

Answer 1

The   neutralization reaction between  potassium  hydroxide  and  sulfuric  acid   is  as  follows
2KOH  +  H2SO4 ---> K2SO4  +  2H2O

number  of   moles  of  KOH=  (43.74  x  0.500)/  1000=  0.02187 moles

the  reacting ratio  of  KOH  to H2SO4  is  2:1  therefore  the   moles  of  H2SO4  is  =  0.021187/2=  0.01094 moles

concentration(molarity) = ( 0.01094/50 ) x 1000=  0.2188M

Answer 2

Answer: The concentration of sulfuric acid is 0.219 M

Explanation:

To calculate the concentration of acid, 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:

$n_1=2\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=43.74mL$

Putting values in above equation, we get:

$2\times M_1\times 50.00=1\times 0.500\times 43.74\\\\M_1=\frac{1\times 0.500\times 43.74}{2\times 50.00}=0.219M$

Hence, the concentration of sulfuric acid is 0.219 M