Answer:
[tex]M_{base}=0.173M[/tex]
Explanation:
Hello there!
In this case, since the titration of acids like KHP with bases like NaOH are performed in a 1:1 mole ratio, it is possible for us to know that their moles are the same at the equivalence point, and the concentration, volume and moles are related as follows:
[tex]n_{acid}=M_{base}V_{base}[/tex]
Thus, by solving for the volume of the base as NaOH, we obtain:
[tex]M_{base}=\frac{n_{acid}}{V_{base}} \\\\M_{base}=\frac{0.8675g*\frac{1mol}{204.22g} }{0.02456L} \\\\M_{base}=0.173M[/tex]
Best regards!
