Question

QUESTION 8

Answer 1

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 $NaOH$.

$1\times 0.500M\times 25ml=1\times M_2\times 23.5ml$

$M_2=0.532M$

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

Answer 2

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