Given:
M(HCl) = 0.62 M
V(HCl) = 16.129 mL
M(B) = 0.25 M
V(B) = 40 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.62 M * 16.129 mL = 10 mmol
mol(B) = M(B) * V(B)
mol(B) = 0.25 M * 40 mL = 10 mmol
We have:
mol(HCl) = 10 mmol
mol(B) = 10 mmol
10 mmol of both will react to form BH+ and H2O
BH+ here is strong acid
BH+ formed = 10 mmol
Volume of Solution = 16.129 + 40 = 56.129 mL
Ka of BH+ = Kw/Kb = 1.0E-14/7.6E-6 = 1.316*10^-9
concentration ofBH+,c = 10 mmol/56.129 mL = 0.1782 M
BH+ + H2O -----> B + H+
0.1782 0 0
0.1782-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 ((1.316*10^-9)*0.1782) = 1.531*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.531*10^-5 M
[H+] = x = 1.531*10^-5 M
use:
pH = -log [H+]
= -log (1.531*10^-5)
= 4.815
Answer: 4.81
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