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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

Verified Answer :

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 : 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

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