Question

A titration is performed to determine the amount of sulfuric acid, H2SO4, in a 6.5 mL sample taken from car battery. About 50 mL of water is added to the sample, and then it is titrated with 43.37 mL of a standard 0.5824 molar NaOH solution. You balanced this reaction in a previous problem. How what is the molar concentration of sulfuric acid in the original sample

Answer 1

Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M

Step-by-step explanation:

To calculate the molarity 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 NaOH.

We are given:

`n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL`

Putting values in above equation, we get:

`2* M_1* 56.5=1* 0.5824* 43.37`

`M_1=0.2235`

Now to calculate the molarity of original solution:

`M_1* 6.5=0.2235* 56.5`

`M_1=1.943`

Thus the molar concentration of sulfuric acid in the original sample is 1.943 M