Question

In a titration, 25.0 mL of 0.500 M KHP is titrated to the equivalence point with NaOH. The final solution volume is 53.5 mL. What is the value of M2 for the reaction?
0.439 M

0.500 M

0.234 M

1.07 M

Answer 1

Answer : The concentration of the NaOH is, 0.234 M

Explanation :

Using neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of an acid = 1

$n_2$ = acidity of a base = 1

$M_1$ = concentration of KHP = 0.500 M

$M_2$ = concentration of NaOH = ?

$V_1$ = volume of KHP = 25 ml

$V_2$ = volume of NaOH = 53.5 ml

Now put all the given values in the above law, we get the concentration of the NaOH.

$1\times 0.500M\times 25ml=1\times M_2\times 53.5ml$

$M_2=0.234M$

Therefore, the concentration of the NaOH is, 0.234 M

Answer 2

M1v1=m2v2 
m2=(m1v1)/v2 
Where m is the molarities and v is the volumes
m2=(25.0*0.500)/53.5
m2=12.5/53.5
m2=0.2336
by rounding off:
m2=0.234 M
so the answer is C: 0.234 M