Calculate pH of a weak acid/conjugate base buffer solution.
1. a) Calculate the pH of 650. mL of a
0.211-M solution of acetic acid
before and after the addition of 0.123 mol of
potassium acetate.
pH befor addition =
pH after addition =
1 b) Calculate pH of a weak base/conjugate acid buffer solution.
A 0.190-M aqueous solution of
C2H5NH2
(ethylamine) has a pH of 11.9.
Calculate the pH of a buffer solution that is
0.190 M in
C2H5NH2 and
0.487 M in
C2H5NH3+.
pH =
1. c) Use the Henderson–Hasselbalch equation to calculate pH of a buffer solution.
Using the Henderson-Hasselbalch equation, calculate the pH of a
buffer solution that is 0.450 M in
HSO3- and
0.204 M in
SO32-.
pH =
1d. ) Calculate buffer pH after adding strong acid or strong base.
Determine the pH change when
0.084 mol HNO3 is
added to 1.00 L of a buffer solution that is
0.489 M in HClO and
0.353 M in
ClO-.
pH after addition − pH before addition = pH change =
1.Calculate the pH of 650. mL of a 0.211-M solution of acetic acid before and after the addition of 0.123 mol of potassium acetate.
pH before addition =
pH after addition =
Ans :-
pH before addition :-
ICE table of the reaction is :
............................CH3COOH (aq)-------------------------> CH3COO- (aq) ...................+...............H+ (aq)
Initial .....................0.211 M...............................................0.0 M...................................................0.0 M
Change .....................-y......................................................+y........................................................+y
Equilibrium ..............(0.211-y) M...........................................y M.....................................................y M
Expression of Ka is :-
Ka = [CH3COO-].[H+] /[CH3COOH]
1.8 x 10-5 = y2(0.211-y)
y2 + 1.8 x 10-5 y - 3.798 x 10-6 = 0
Therefore ,
y = 0.00194 M
Therefore,
[H+] = 0.00194 m
pH = - log [H+]
pH = - log 0.00194 M
pH = 2.71
------------------------------------------
pH after addition :
Moles of acetic acid = Molarity x Volume of solution
= 0.211 M x 0.650 L
= 0.13715 mol
Given moles of CH3COOK = 0.123 mol
By using Henderson- Hasselbalch equation :
pH = pKa + log [CH3COOH]/[CH3COOK]
pH = 4.74 + log 0.123 mol / 0.13715 mol
pH = 4.74 - 0.0473
pH = 4.70
Calculate pH of a weak acid/conjugate base buffer solution. 1. a) Calculate the pH of 650....
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