A buffer is prepared by adding 21.0g of sodium acetate (CH3COONa) to 500mL of a
ID: 854112 • Letter: A
Question
A buffer is prepared by adding 21.0g of sodium acetate (CH3COONa) to 500mL of a 0.145M acetic acid (CH3COOH) solution.
Part A Determine the pH of the buffer.
Express your answer using two decimal places.
pH=________
Part B Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Part C Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Explanation / Answer
Molar mass CH3COONa = 82.03g/mol
moles in 21.0g = 21.0/82.03 = 0.256 mol
Molarity of CH3COONa in solution = 0.256/0.500 = 0.512M
pKa CH3COOH = 4.74
Use the Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.512/0.155)
pH = 4.74 + 0.519
pH = 4.74 + 0.519
pH = 5.259
.
Adding HCl to the buffer: Adding HCl adds an increase in H+ ions and the pH should go down
When you add the HCl to the buffer there is no reaction with the CH3COOH - acids do not react with acids .
The HCl reacts with the dissociated CH3COO- from the CH3COONa in solution . This produces more CH3COOH a,d Cl- ions
You can represent this in an ionic equation:
CH3COO-(aq) + Na+(aq) + H+(aq) + Cl-(aq) ? CH3COOH (l) +Na+(aq) + Cl-(aq)
And as a NET ionic equation:
CH3COO-(aq) + H+(aq) ? CH3COOH(l)
The CH3COOH is shown as (l) because it is un-ionised in the solution.
Basically this is what happens in a buffer - you add HCl - the increased H+ ions are reacted by the CH3COO- ( called the conjugate base) to produce un - ionised CH3COOH . Because you have removed the excess H+ ions the pH of the buffer does not change a lot on the addition of HCl.
Adding NaOH to the buffer:
In adding NaOH you are adding an increased quantity of OH- ions. The pH should go up .
The most likely substance that these hydroxide ions are going to collide with is an acetic acid , CH3COOH in solution.
CH3COOH (l) + NaOH(aq) ? CH3COONa(aq) + H2O(l)
The ionic equation:
CH3COOH (l) + Na+(aq) + OH-(aq) ? CH3COO- (aq) + Na+(aq) + H2O(l)
And the NET ionic equation
CH3COOH(l) + OH- (aq) ? CH3COO- (aq) + H2O(l)
The reaction produces acetate ions and water.
In this way most of the new hydroxide ions are removed and the pH doesn't increase very much.
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