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.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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
x2 = 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
Calculate the pH of a 0.10 M solution of sodium acetate
Since base is always equal to COAc > Kb , neglect x compared to COAc. Then,
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.
Calculate the pH of a buffer prepared by adding
10 mL of 0.10 M acetic acid to 20 mL of 0.10 M
0.10 mmol/mL X 10 mL = MHOAc X 30 mL
MHOAc = 0.033 mmol/mL
For OAcˉ , solution. The final volume is 30
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.
HA H⁺ + A⁻
β = dCBOH/ dpH
= - dCHA / dpH
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.
Solution : mol/L, the change in pH is :
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.
Since amount of strong acid or base greater than the equilibrium amount of H⁺ or OHˉ in the buffer. pOH =pKw -pH, we can also write , the above equation form pKw
Example : 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
We want 100 amount of strong acid or base greater than the equilibrium amount of H⁺ or OHˉ in the buffer. mL 0f 0.200 M NH4Cl.
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
The molarity of concentrated ammonia is 14.8 M amount of strong acid or base greater than the equilibrium amount of H⁺ or OHˉ in the buffer.