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wlad13 [49]
3 years ago
15

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. Wha

t is the value of M2 for the reaction?
0.439 M

0.500 M

0.234 M

1.07 M
Chemistry
2 answers:
kumpel [21]3 years ago
8 0

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\times 0.500M\times 25ml=1\times M_2\times 53.5ml

M_2=0.234M

Therefore, the concentration of the NaOH is, 0.234 M

Lena [83]3 years ago
3 0
M1v1=m2v2 
m2=(m1v1)/v2 
Where m is the molarities and v is the volumes
<span>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</span>
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