2)
NH3 dissociates as:
NH3 +H2O -----> NH4+ + OH-
0.1 0 0
0.1-x x x
Kb = [NH4+][OH-]/[NH3]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.8*10^-5)*0.1) = 1.342*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
1.8*10^-5 = x^2/(0.1-x)
1.8*10^-6 - 1.8*10^-5 *x = x^2
x^2 + 1.8*10^-5 *x-1.8*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-5
c = -1.8*10^-6
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 7.2*10^-6
roots are :
x = 1.333*10^-3 and x = -1.351*10^-3
since x can't be negative, the possible value of x is
x = 1.333*10^-3
So, [OH-] = x = 1.333*10^-3 M
use:
pOH = -log [OH-]
= -log (1.333*10^-3)
= 2.87
use:
PH = 14 - pOH
= 14 - 2.87
= 11.13
Answer: 11.13
3)
Concentration after mixing = mol of component / (total volume)
M(CHO2-) after mixing = M(CHO2-)*V(CHO2-)/(total volume)
M(CHO2-) after mixing = 0.2 M*100.0 mL/(100.0+250.0)mL
M(CHO2-) after mixing = 5.714*10^-2 M
Concentration after mixing = mol of component / (total volume)
M(HCHO2) after mixing = M(HCHO2)*V(HCHO2)/(total volume)
M(HCHO2) after mixing = 0.15 M*250.0 mL/(250.0+100.0)mL
M(HCHO2) after mixing = 0.1071 M
Ka = 1.8*10^-4
pKa = - log (Ka)
= - log(1.8*10^-4)
= 3.745
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.745+ log {5.714*10^-2/0.1071}
= 3.472
Answer: 3.47
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