A 4.00*10^2 mL solution is 0.25 M in ethylamine (CH3CH2)NH2 and 0.40 M in ethyl
ID: 590422 • Letter: A
Question
A 4.00*10^2 mL solution is 0.25 M in ethylamine (CH3CH2)NH2 and 0.40 M in ethyl ammonium chlorate (CH3CH)NH3ClO3 a) what is the pH of solution b) what is the pH of the solution after adding 125mL of 0.64 M of HBr? c) what is the pH of the solution after adding 0.050 mol of KOH to the original 400mL solution? A 4.00*10^2 mL solution is 0.25 M in ethylamine (CH3CH2)NH2 and 0.40 M in ethyl ammonium chlorate (CH3CH)NH3ClO3 a) what is the pH of solution b) what is the pH of the solution after adding 125mL of 0.64 M of HBr? c) what is the pH of the solution after adding 0.050 mol of KOH to the original 400mL solution? a) what is the pH of solution b) what is the pH of the solution after adding 125mL of 0.64 M of HBr? c) what is the pH of the solution after adding 0.050 mol of KOH to the original 400mL solution?Explanation / Answer
a) Henderson equation is
pOH = pKb + log([BH+]/[B])
= pKb + log ( [ CH3CH2NH3+]/[CH3CH2NH2] )
= 3.2 + log ( 0.40/0.25 )
= 3.2 + 0.20
= 3.4
pH = 14 - pOH
= 14 - 3.4
= 10.6
b) HBr react with CH3CH2NH2
CH3CH2NH2 + HBr -------> CH3CH2NH3+
This reaction is 1:1 reaction
no of mole of HBr added =( 0.64mol/1000ml)*125ml = 0.08mol
initial mole of CH3CH2NH2 = (0.25mol/1000ml)*400ml = 0.1mol
remaining mole of CH3CH2NH2 = 0.1-0.08 = 0.02
initial mole of CH3CH2NH3+ = (0.40/1000ml)*400ml =0.16
mole of CH3CH2NH3+ after addition = 0.16 + 0.08 = 0.24
final voleume = 525ml
[CH3CH2NH2] =( 0.02mol/525ml)*1000ml =0.0381M
[CH3CH2NH3+] = (0.24mol/525ml)*1000ml = 0.4571M
pOH = pKa + log( 0.4571/0.0381)
= 3.2 + 1.08
= 4.28
pH = 14- 4.28
= 9.72
c) KOH react with conjucate acid CH3CH2NH3+
CH3CH2NH3+ + OH- -------> CH3CH2NH2 + H2O
Mole of NaoH added = 0.05
after addition
mole of CH3CH2NH3+= 0.16 -0.05 = 0.11
mole of CH3CH2NH2 = 0.10 + 0.05 = 0.15
[ CH3CH2NH2] = (0.15mol/400ml) *1000 ml = 0.375M
[ CH3CH2NH3+] = (0.11mol/400ml)*1000ml = 0.275M
pOH = 3.2 + log(0.275/0.375)
= 3.2 - 0.13
= 3.07
pH = 14 - 3.07
= 10.93
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