Question

QUESTION 8

In a titration, 25.0 mL of 0.500 M KHP is titrated to the equivalence point with NaOH. The final solution volume is 48.5 mL. What is the value of M2 for the reaction?

0.256 M

0.532 M

0.970 M

0.500 M

Answer 1

m1v1=m2v2
m2=(m1v1)/v2
Where m is the molarities and v is the volumes
m2=(0.500*25.0)/48.5
m2=12.5/48.5
m2=0.256 M
so it will be A: 0.256 M

Answer 2

Answer : The value of
`M_2` (concentration of NaOH) for the reaction will be, 0.532 M

Explanation :

Using neutralization law,

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1` = basicity of an acid (KHP) = 1

`n_2` = acidity of a base (NaOH) = 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 = 48.5 - 25 = 23.5 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* 23.5ml`

`M_2=0.532M`

Therefore, the value of
`M_2` (concentration of NaOH) for the reaction will be, 0.532 M