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

Answer: The concentration of sulfuric acid is 0.219 M

Step-by-step 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* M_1* 50.00=1* 0.500* 43.74\\\\M_1=(1* 0.500* 43.74)/(2* 50.00)=0.219M`

Hence, the concentration of sulfuric acid is 0.219 M

Answer 2

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