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

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

Answer 2

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* 0.500M* 25ml=1* M_2* 53.5ml`

`M_2=0.234M`

Therefore, the concentration of the NaOH is, 0.234 M