F. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum b
ID: 880355 • Letter: F
Question
F. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that ''one solution is a buffer'' is not sufficient. Ka of acetic acid = 1.8 x 10^-5 Acetic acid/sodium acetate buffer: Acetic acid = 0.1M, Sodium acetate = 0.1MExplanation / Answer
Given
Volume of Buffer is 10.0 mL , volume of water = 10.0 mL
We add 0.25 mL , 0.10 M HCl to both of the solution.
Lets calculate moles of HCl
Mol of HCl = Volume of HCl in L * molarity
= 0.23 E-3 L * 0.10 M
= 2.5E-5 mol HCl
Lets calculated concentration of HCl in water
Total volume of solution = 0.010 L + 0.25E-3 L
= 0.01025 L
Molarity of HCl = 2.5E-5 mol / 0.01025 L = 0.002439 M
We know HCl is strong acid and so it dissociates fully in water.
[H3O+] = [HCl]
So [H3O+ ] = 0.002439 M
pH = -log [H3O+] = - log 0.002439 =2.613
Concentration of H3O+ in buffer.
We know HCl reacts with acetate ions and forms acetic acid.
According to the reaction 1 mol HCl gives 1 mol acetic acid and same time one mol of acetate ion is reduced.
We use this argument in Henderson equation to get pH
pH = pka + log ([base]/[acid])
base is conjugate base of the acid.
Pka = - log ka
Ka of acetic acid = 1.75E-5
Pka = - log 1.75E-5 = 4.76
Lets assume in the given buffer moles of each base and acid are same.
That is 1 mol each.
pH = 4.76 + log ( [1 – 2.5E-5 ]/[1+ 2.5 E-5)
= 4.76
[H3O+] = antilog (-pH )
= Antilog (-4.76)
= 1.75E-5 M
pH is un altered or we say remains same.
If we add the HCl to the water then its pH is changed from neutral to 2.613
But if add the HCl to buffer then its pH is not changed.
When HCl reacts with base and pH increase by some unit, Same time it produces acid and then pH decrease by same unit.
We write equation to know this.
HCl + CH3COO- --- > CH3COOH + Cl-
This clearly shows that one mole base is reduced and same time 1 mol of acid is formed.
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