Loading in 2 Seconds...
Loading in 2 Seconds...
Weak acids and bases Salt of weak acid and bases buffer Lecture 9 9 Feb 2011. Noorulnajwa Diyana Yaacob firstname.lastname@example.org. Weak Acid. A weak acid is an acid that does not completely donate all of its hydrogens when dissolved in water.
Calculate the pH and pOH of a
1.0 x 10-3 solution of acetic acid
HOAc H+ + OAc-
Ka= [H+][OAc-] = 1.75 x 10-5
1.0 x 10-3 - x
x is smaller than C, neglect it, therefore,
x2 = 1.75 x 10-5 = 1.32 x 10-4M = [H+ ]
1.0 x 10-3
pH = -log 1.32 x 10-4 = 3.88
pOH = 14.00 – 3.88 = 10.12
Ka =[H+][OAc-] = 1.75 x 10-5
The value of Kb can be calculated from Ka of acetic acid and Kw, if we multiply both the numerator and denominator by [H+]:
The product of Ka of any weak acid and Kb of its conjugate base is always equal to Kw:
The pH of such a salt (Bronstead base) is calculated as the same manner as for any other weak base When the salts hydrolyzes, it forms an equal amount of HA and OH- ,If the  of A- is CA-, then,
Calculate the pH of a 0.10 M solution of sodium acetate
Buffer solution consists mixture of weak acid and its conjugate base or weak base with it conjugate acid at predetermined concentration or ratios.
CH3COOH (aq) CH3COO‒ (aq) + H+ (aq)
It is known as an acidic buffer solution and it maintains a pH value that is less than 7.
NH3 (aq) + H2O(l) NH4+ (aq) + OH‒ (aq)
It is known as a basic buffer solution and it maintains a pH value that is greater than 7.
This form of the ionization constant equation is called the Handerson-Hasselbalch equation
Calculate the pH of a buffer prepared by adding
10 mL of 0.10 M acetic acid to 20 mL of 0.10 M
Need to calculate the [ ] of the acid and salt in the solution. The final volume is 30 mL:
0.10 mmol/mL X 10 mL = MHOAc X 30 mL
MHOAc = 0.033 mmol/mL
0.10 mmol/mL X 20 mL = MOAcˉ x 30 mL
MOAcˉ = 0.067 mmol/mL
= 4.76 + log 2.0
pH = constant + log [A⁻]
*if solution is diluted the ratio remains constant
So, the pH of the solution does not change.
If small amount of strong acid added it will combined with an equal amount of the A⁻ to convert it to HA.
HA H⁺ + A⁻
If small amount of strong base is added it will combined with part of HA to form an equivalent amount of A⁻. Again, change in ratio is small.
β = dCBOH/ dpH
= - dCHA / dpH
The CHA and CAˉ represent the analytical [ ] of the acid and its salt respectively.
If we add solid sodium hydroxide until it becomes 0.0050 mol/L, the change in pH is :
A buffer solution is 0.20 M in acetic acid and
in sodium acetate. Calculate the change in pH
upon adding 1.0 mL of 0.10M hydrochloric acid
to 10 mL of this solution.
mmol HOAc = 2.0 + 0.1 = 2.1 mmol
mmol OAcˉ = 2.0 – 0.1 = 1.9 mmol
The change in pH is -0.05.
A buffer can resist a pH change even when there is added an amount of strong acid or base greater than the equilibrium amount of H⁺ or OHˉ in the buffer.
Calculate the volume of concentrated ammonia
and the weight of ammonium chloride you
would have to take to prepare 100 mL of a
buffer at pH 10.00 if the final concentration of
salt is said to be 0.200 M
mmol NH4Cl = 0.200 mmol/mL × 100 mL
= 20.0 mmol
mg NH4Cl = 20.0 mmol × 53.5 mg/mmol
= 1.07 × 10³ mg
So, 1.07 g NH4Cl. Calculate [ ] of NH3 by