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

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. 

Answer 2

Answer: The concentration of $H_2SO_4$ comes out to be 0.16 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 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:

$2\times M_1\times 20.5=1\times 0.15\times 43.2\\\\M_1=0.16M$

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