Buffer capacity refers to the amount of acid or base a buffer can absorb without
ID: 497199 • Letter: B
Question
Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times a much base a 0.1 M buffer for a given pH change. In this problem begin with buffer of known pH and concentration and calculate the new pH after a particular quantity acid or base is of added. You are given 60 mL of 0.50 M phosphate buffer, pH = 6.83, to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: Concentration of HPO^2_4: 0.304 M concentration of H_2PO_4: 0.196 M Molar quantity of HPO^2_4: 18.2 mmol Molar quantity of H_2PO_4: 11.8 mmol You add 1.7 mL of 1.00 M HCI to the buffer. Calculate the molar quantity of H_3O added as HCI, and the final molar quantity of HPO^2-_4. and H_2PO_4 at equilibrium? What is the new HPO^2-_4/H_2PO_4- ratio, and the new pH of the solution? The pK_a of H_2PO_4 is 6.64. Use the Henderson-Hasselbach equation to calculate the new pH. Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NaOH. Using steps similar to those above, calculate the new pH of the solution.Explanation / Answer
Buffer solution resists change in pH with dilution or with addition of acid or base.
Given is phosphate buffer
H2PO4- + H2O ----> HPO42- + H3O+
HPO42- + H2O------> H2PO4- + OH-
To find pH of buffer we express the equilibrium concentrations in terms of analytical concentrations
First reaction decreases the concentration of H2PO4- by amount equal to [ H3O+] and second reaction increases the concentration in terms of [OH-]
[H2PO4-] = cH2PO4- - [ H3O+] + [OH-]
Simillarly first reaction increases the concentration of HPO42- in terms of [ H3O+] and second reaction decreases the concentration by amount equal to [OH-]
[HPO42-] = cHPO2-4 + H3O+] - [OH-]
The difference in concentration [ H3O+] and [OH-] is very small compared to the molar concentration acid and conjugate base.
[H2PO4-] ~ cH2PO-4 ----- 1
[HPO42-] ~ cHPO2-4 ----- 2
we know the Ka = [HPO42-] [ H3O+] / [H2PO4-]
on substituting
[ H3O+] = Ka cH2PO-4 / cHPO2-4
a)
Effect of added HCl
Given 60ml of 0.50M buffer
[H2PO4-] = 0.304M
[HPO42-]= 0.196M
Now we add 1.7ml of 1.00M HCl
The added acid converts parts of HPO42- to H2PO4-
HPO42- + H3O+ ----> H2PO4- + H2O
From 1 and 2 it is clear that analytical concentration is equal to equilibrium concentration
On adding HCl the concentration of HPO42- is decreased
Initial number of moles of HPO42- is
= 60 mL x 0.196 mM/mL
= 0.01176 mol
Number of moles of HCl added is
= 1.7 mL x 1.00 mM/mL
= 0.0017 mol
Number of moles of HPO42- after adding HCl is
= (0.01176 - 0.0017 )mol
Number of moles of HPO42- after adding HCl = 0.01006 mol
Concentration of HPO42 after adding HCl is = 0.01006 mol / (60+1.7) => 1.63 x 10-4 M
Like wise On adding HCl the concentration of H2PO4- is increased.
Initial no.of moles of H2PO4- = 0.304 mmol /mL x 60 ml => 0.01824 mol
Number of moles of H2PO4- after adding HCl (0.01824 + 0.0017 ) is 0.01994 mol
Concentration (0.01994 / 61.7) = 3.23 x 10-4
To find the concentration of [H3O+] we use
[H3O+] = Ka cH2PO-4 / cHPO2-4
[H3O+] = Ka 3.23 x 10-4 / 1.63 x 10-4
pKa is given in the part b of the question as 6.64
[H3O+] = 10-6.64 (3.23 x 10-4 / 1.63 x 10-4)
[H3O+] = 4.539 x 10-7
b)
The ration of new HPO42- / H2PO4- is
= 0.01006 / 0.01994
= 0.50 (mole ratio)
The Henderson-Hasselbalch equation is
pH = pKa + log (cA- / cHA)
From the above we have
Number of moles of HPO42- = 0.01006 mol
Number of moles of H2PO4- = 0.01994 mol
pka = 6.64
pH = 6.64 + log(0.01006 / 0.01994) (as instructed we are using mole ratio )
pH = 6.34
c)
On adding NaOH some of H2PO4- is converted to HPO42-
H2PO4- decreases
Number of moles of H2PO4- = 0.01824 mol - (3.7mL x 1.00 mmol/mL)
= 0.01454 mol
HPO42- increases
Number of moles of HPO42- = 0.01176 mol + (3.7mL x 1.00 mmol/mL)
= 0.01546 mol
pH = pKa + log (cA- / cHA)
pH = 6.64 + log(0.01546 / 0.01454)
pH = 6.66
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