Question

A titration reached the equivalence point when 17.0 mL of was added to of NaOH (aq) of unknown concentration. What is the concentration (M) of this unknown NaOH solution? H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

Answer 1

This is an incomplete question, here is a complete question.

A titration reached the equivalence point when 17.0 mL of 0.211 M H₂SO₄ (aq) was added to 16.3 mL of NaOH (aq) of unknown concentration. What is the concentration (M) of this unknown NaOH solution?

`H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)`

Answer : The concentration (M) of this unknown NaOH solution is, 0.440 M

Explanation :

To calculate the concentration of unknown NaOH solution, 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=0.211M\\V_1=17.0mL\\n_2=1\\M_2=?\\V_2=16.3mL`

Now put all the given values in above equation, we get:

`2* 0.211M* 17.0mL=1* M_2* 16.3mL`

`M_2=0.440M`

Therefore, the concentration (M) of this unknown NaOH solution is, 0.440 M