Answer : The final concentration is, 0.01 mole/L or 0.01 M
Solution :
According to the neutralization law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = initial molarity (concentration) of solution = 0.04 M = 0.04 mole/L
[tex]V_1[/tex] = initial volume of solution = 50 ml = 0.05 L
[tex]M_2[/tex] = final molarity (concentration) of solution = ?
[tex]V_2[/tex] = final volume of solution = 200 ml = 0.2 L
Conversion : (1 L = 1000 ml)
Now put all the given values in the above law, we get the volume of NaOH solution.
[tex](0.04mole/L)\times (0.05L)=(M_2)\times (0.2L)[/tex]
[tex]M_2=0.01mole/L=0.01M[/tex]
Therefore, the final concentration is, 0.01 mole/L or 0.01 M
