(competency) Two buffers are prepared by adding an equal number of moles of form
ID: 963598 • Letter: #
Question
(competency) Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make 1.00 L of solution. Buffer A prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. Calculate the pH of each buffer, and explain why they are equal. Which buffer will have the greater buffer capacity? Explain. Calculate the change in pH for each buffer upon the addition of 1.0 mL of 1.00 M HCI. Calculate the change in pH for each buffer upon the addition of 10 mL of 1.00 M HCI. Discuss your answers for parts c and d in light of your response to part b.Explanation / Answer
pH = pKa + logBase/acid)
For buffer A [base ] = 1.00 and [Acid ]= 1.00
pH = pKa + logBase/acid)
pH = 3.77 + log(1.00/1.00)
= 3.77 + 0 = 3.77
For buffer B [base ] = 0.010 and [Acid ]= 0.010
pH = pKa + logBase/acid)
pH = 3.77 + log(0.010/0.010)
pH = 3.77 + log 1
= 3.77 + 0 = 3.77
pH of both buffer are same because the moles of acid and base are same.
b. Buffer solution is able to retain almost constant pH when small amount of acid/base is added.
The buffer capacity is the quantitative measure of this resistance to pH changes . he buffer capacity is called as the amount of strong acid or base, in gram-equivalents, that must be added to 1 liter of the solution to change its pH by one unit. Thus the buffer A has greater buffer capacity.
c
1 M buffer, 1 mL HCl)
For buffer A
1 mL of 1 M HCl will be 10^-3 M HCl.
pH = 3.77 + log(1.00 – 0.001)/(1.00+0.001)
pH = 3.77 – 0.000434
pH = 3.7695
For buffer B
1 mL of 1 M HCl will be 10^-3 M HCl.
pH = 3.77 + log(0.010 – 0.001)/(0.010+0.001)
pH = 3.77 + log 0.009/0.011
pH = 3.77 – 0.087
pH = 3.683
d.
1 M buffer, 10 mL HCl)
10 mL of 1 M HCl will be 10^-2 M HCl (without the small dilution effect by adding 10 mL to 1 L)
For buffer A
pH = 3.77 + log (1.00 – 0.02)/(1.00 + 0.02)
pH = 3.77 – 0.01737
pH = 3.7526
For buffer B
pH = 3.77 + log (0.010 – 0.02)/(0.010 + 0.02)
pH = 3.77 -
means pH of this condition not to calculate with Henderson-Hasselbalch equation.
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