Question

A titration was performed in a lab situation. H2SO4 was titrated with NaOH. The following data was collected: mL of NaOH used = 43.2 mL concentration NaOH = 0.15 M mL H2SO4 = 20.5 mL Notice that H2SO4 releases 2 H+ per mole. What is the concentration of H2SO4? 0.036 M 0.16 M 0.63 M 6.3 M

Answer 1

Answer: The concentration of
`H_2SO_4` comes out to be 0.16 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 NaOH.

We are given:

`n_1=2\\M_1=?M\\V_1=20.5mL\\n_2=1\\M_2=0.15M\\V_2=43.2mL`

Putting values in above equation, we get:

`2* M_1* 20.5=1* 0.15* 43.2\\\\M_1=0.16M`

Hence, the concentration of
`H_2SO_4` comes out to be 0.16 M.

Answer 2

For the titration we use the equation,
M₁V₁ = M₂V₂
where M is molarity and V is volume. Substituting the known values,
(0.15 M)(43.2 mL) = (2)(M₂)(20.5 mL)
We multiply the right term by 2 because of the number of H+ in H2SO4. Calculating for M₂ will give us 0.158 M. Thus, the answer is approximately 0.16M.