b. 50.0 mL of 0.200 M HBr
a)
find the volume of HBr used to reach equivalence point
M(B)*V(B) =M(HBr)*V(HBr)
0.1 M *50.0 mL = 0.2M *V(HBr)
V(HBr) = 25 mL
Given:
M(HBr) = 0.2 M
V(HBr) = 25 mL
M(B) = 0.1 M
V(B) = 50 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.2 M * 25 mL = 5 mmol
mol(B) = M(B) * V(B)
mol(B) = 0.1 M * 50 mL = 5 mmol
We have:
mol(HBr) = 5 mmol
mol(B) = 5 mmol
5 mmol of both will react to form BH+ and H2O
BH+ here is strong acid
BH+ formed = 5 mmol
Volume of Solution = 25 + 50 = 75 mL
Ka of BH+ = Kw/Kb = 1.0E-14/7.0E-9 = 1.429*10^-6
concentration ofBH+,c = 5 mmol/75 mL = 0.0667 M
BH+ + H2O -----> B + H+
6.667*10^-2 0 0
6.667*10^-2-x x x
Ka = [H+][B]/[BH+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.429*10^-6)*6.667*10^-2) = 3.086*10^-4
since c is much greater than x, our assumption is correct
so, x = 3.086*10^-4 M
[H+] = x = 3.086*10^-4 M
use:
pH = -log [H+]
= -log (3.086*10^-4)
= 3.5106
Answer: 3.51
b)
Given:
M(HBr) = 0.2 M
V(HBr) = 50 mL
M(B) = 0.1 M
V(B) = 50 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.2 M * 50 mL = 10 mmol
mol(B) = M(B) * V(B)
mol(B) = 0.1 M * 50 mL = 5 mmol
We have:
mol(HBr) = 10 mmol
mol(B) = 5 mmol
5 mmol of both will react
excess HBr remaining = 5 mmol
Volume of Solution = 50 + 50 = 100 mL
[H+] = 5 mmol/100 mL = 0.05 M
use:
pH = -log [H+]
= -log (5*10^-2)
= 1.301
Answer: 1.30
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