Please answer the entire question and show all work. 3. A buffer is a solution c
ID: 576658 • Letter: P
Question
Please answer the entire question and show all work.
3. A buffer is a solution containing a mixture of an acid and its conjugate base. Buffers maintain a constant pH even if H' is added or removed by chemical processes. With that in mind a. Suppose that we wanted to make a buffer that would maintain a constant pH of 10.3 using bicarbonate (HCO3) and carbonate (CO3 species. The pK, for the conversion of bicarbonate to carbonate is 10.3. What ratio of carbonate:bicarbonate would the buffer need to contain in order to have this pH? b. How much sodium carbonate and sodium bicarbonate (in moles) would we need to use to make 500 mL of this buffer, assuming we wanted the sum of [COHCOsin the buffer to equal 20 mM? c. Carry out the same calculations for a bicarbonate/carbonate buffer at a pH of 9.9. What ratio of carbonate:bicarbonate would the buffer need to contain in order to have this pH? Assuming we wanted the sum of (CO[HCOs] in the pH 9.9 buffer to equal 20 mM, what concentrations of would we need to achieve the correct ratio? d. Now suppose that we wanted to make a buffer that would maintain a constant pH of 5.0 using bicarbonate and carbonate. Again calculate the ratio of (CO:(HCO.Assuming we wanted the sum of [CO HCOs ] in the pH 5.0 buffer to equal 20 mM, what concentrations would we need to achieve the correct ratio?Explanation / Answer
a) According to the Henderson- Hasselbalch equation,
pH= pKa+ log [A-]/[HA]
=> 10.3=10.3+ log [A-]/[HA]
=> 0= log[A-]/[HA]
=> [A-]/[HA]=1
From the above equation we can conclude that we will need equal amounts of carbonate and bicarbonate or in other words the ratio of carbonate and bicarbonate should be 1:1
b) Since we know that the quantities of Carbonate and Bicarbonate are in the ratio 1:1
We can say Carbonate and Bicarbonate both equals 20 nM
=> Sodium carbonate required= (0.020 mol/L X 0.5L)* 105.99 g/mol = 1.0599 grams
=> Sodium bicarbonbate required= (0.020 mol/L X 0.5L)* 84.01 g/mol = 0.8401 grams
c) using the Henderson-Hasselbalch equation again,
pH= pKa+ log[A-]/[HA]
9.9= 10.3+ log[A-]/[HA]
log[A-]/[HA]= -0.4
[A-]/[HA]= 0.398
So the ratio should be 0.398:1 which should be approximately 2:5 in standard form