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?
[tex]H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)[/tex]
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:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=0.211M\\V_1=17.0mL\\n_2=1\\M_2=?\\V_2=16.3mL[/tex]
Now put all the given values in above equation, we get:
[tex]2\times 0.211M\times 17.0mL=1\times M_2\times 16.3mL[/tex]
[tex]M_2=0.440M[/tex]
Therefore, the concentration (M) of this unknown NaOH solution is, 0.440 M
