If 24.83 cm3 of 0.15 M NaOH is titrated to its end point with 39.45 cm3
of HCl, what is the molarity of the HCl?

a

0.094 M

b

0.150 M

c

0.940 M

d

1.500 M

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a

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Discussions (5)

Ajanaku Samiat
11 years ago

Ma*Va/Mb*Vb=Na/Nb

Ma*39.45/0.15*24.83=1/1

Ma=0.094M, the correct ans is A

Tumise17
1 year ago

M1.V1=M2.V2
0.15*24.83=39.45*M2
M2=0.15*24.83/39.45
=0.0944M

Ibto
5 years ago

how do we know the ratio

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