Respuesta :
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
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 : The value of [tex]M_2[/tex] (concentration of NaOH) for the reaction will be, 0.532 M
Explanation :
Using neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity of an acid (KHP) = 1
[tex]n_2[/tex] = acidity of a base (NaOH) = 1
[tex]M_1[/tex] = concentration of [tex]KHP[/tex] = 0.500 M
[tex]M_2[/tex] = concentration of NaOH = ?
[tex]V_1[/tex] = volume of [tex]KHP[/tex] = 25 ml
[tex]V_2[/tex] = 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 [tex]NaOH[/tex].
[tex]1\times 0.500M\times 25ml=1\times M_2\times 23.5ml[/tex]
[tex]M_2=0.532M[/tex]
Therefore, the value of [tex]M_2[/tex] (concentration of NaOH) for the reaction will be, 0.532 M
