Calculate the pH of a strong acid solution formed by mixing 83.3
mL of 0.0297 M HBrO4 with 77.2 mL of 0.129 M
HNO3.
a) The pH of this solution is 0.8.
b) The pH of this solution is 1.1.
c) The pH of this solution is 0.5.
d) The pH of this solution is 4.4.
e) The pH of this solution is 1.9.
Concentration of mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
where C1 --> Concentration of 1 component
V1-->volume of 1 component
C2 --> Concentration of other component
V2-->volume of other component
n1 --> number of particle from 1 molecule of 1st component
n1 = 1 as one molecule of HBrO4 has one H+ ion
n2 --> number of particle from 1 molecule of 2nd component
n2 = 1 as one molecule of HNO3 has one H+ ion
use:
C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
C = (1*0.0297*83.3+1*0.129*77.2)/(83.3+77.2)
C = 0.0775 M
So,
[H+] = 0.0775 M
use:
pH = -log [H+]
= -log (7.75*10^-2)
= 1.11
Answer: b
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