You have 15.00 mL of a 0.150 M aqueous solution of the weak base CH_3 NH_2 (K_b
ID: 531455 • Letter: Y
Question
You have 15.00 mL of a 0.150 M aqueous solution of the weak base CH_3 NH_2 (K_b = 5.00 times 10^-4). This solution will be titrated with 0.150 M HCl. (a) How many mL of acid must be added to reach the equivalence point? (b) What is the pH of the solution before any acid is added? (c) What is the pH of the solution after 5.00 mL of acid has been added? (d) What is the pH of the solution at the equivalence point of the titration? (e) What is the pH of the solution when 20.00 mL of acid has been added?Explanation / Answer
a) CH3NH2 + HCl -------> CH3NH3+ + Cl-
1 mole of HCl react with 1mole of CH3NH2
Therefore, 15ml of 0.150 HCl must be added for equivalence point
b) CH3NH2 + H2O ------> CH3NH3+ + OH-
Kb = [ CH3NH3+ ] [ OH- ] /[ CH3NH2]
at equillibrium [ CH3NH2 ] = 0.150-X
[ OH- ] = X
[ CH3NH3+ ] = X
Kb = 5.00× 10^-4
5.00× 10^-4 = X^2/(0.150 - X)
X^2 + 5.00× 10^-4X - 7.5 × 10^-5 = 0
X = 8.66×10^-3
OH- = 8.66 × 10^-3M
pOH = 2.06
pH = 11.94
c) No of mol in 5ml of HCl solution = (0.150/1000)×5 = 0.00075
0.00075mol react with 0.00075mol of CH3NH2
No of mol of CH3NH3+ produced = 0.00075mol
No of mol of CH3NH2 = 0.00225 - 0.00075 = 0.0015
Total volume = 20ml
[CH3NH2] = (0.0015/20)×1000=0.075M
[ CH3NH3+ ] = (0.00075/20)×1000= 0.0375M
pOH = pKb + log( [ BH+]/[B] )
= 3.30 + log ( ( 0.00075)/(0.0015))
= 3.30 + 0.3010
= 3.60
pH = 10.40
d) At Equivalence point
[ CH3NH3+ ] = 0.150/2= 0.075M
Ka = Kw/Kb
= 1×10^-14/ 5 × 10^-4
= 2.0 × 10^-11
CH3NH3+ + H2O -------> H3O+ + CH3NH2
2.0×10^-11 = [ H3O+ ] [ CH3NH2]/[ CH3NH3+ ]
at equillibrium
[ H3O+ ] = X
[CH3NH2 ] = X
[ CH3NH3+ ] = 0.075 - X
We can assume
0.075 - X = 0.075
X^2 = 1.5×10^-12
X = 1.22×10^-6
[ H3O+ ] = 1.22× 10^-6M
pH = 5.91
e) Out of 20 ml HCl 15 ml go to neutralisation, 5 ml contribute pH
Total volume = 15ml + 20ml = 35ml
No of mol of HCl =( 0.150 mol/1000ml)× 5ml = 0.00075mol
[H3O+ ] = (0.00075mol/35ml)×1000ml = 0.0214M
pH = 1.67
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