Please I need some calculations for part 1 and 2 from this lab! I\'m not sure ho
ID: 1020079 • Letter: P
Question
Please I need some calculations for part 1 and 2 from this lab! I'm not sure how to do it, and I really don't understand what to calculate for this lab! We will be using sodium bicarbonate and the desired pH is 6.07.
1. In advance, write a recipe for making 250.00 mL of a buffer at the pH you have been assigned. The recipe should include detailed, step-by-step procedures that could be replicated by another CHEM&163 student.
2. Calculate the capacity of 25.00mL your buffer to buffer acid. (how many mL of 0.10M HCl?). Calculate the capacity of 25.00mL of your buffer to buffer base. (how many mL of 0.10M NaOH).
Explanation / Answer
The data seems insufficient. However, I shall try to answer as best as possible.
The pH of the buffer system is 6.07. The buffer system is actually a combination of a weak acid and its conjugate salt. Here the weak acid will be carbonic acid and the salt is sodium bicarbonate, H2CO3 and NaHCO3. We shall designate the weak acid as HA and the conjugate salt as A- for simplicity. A weak acid undergoes partial ionization in water.
HA <=======> H+ + A-
The pH of the solution is given by the Henderson-Hasslebach equation as
pH = pKa + log10[A-]/[HA] where Ka is the acid-dissociation constant of the weak acid [Ka = 4.5*10-7 for carbonic acid; pKa = -log10Ka = -log10(4.5*10-7) = 6.3468]
Therefore,
6.07 = 6.3468 + log10[A-]/[HA]
====> log10[A-]/[HA] = -0.2768
====> [A-]/[HA] = 10-0.2768 = 0.52869
Since [A-]/[HA] = 0.52689, we can say [A-]/[HA] =0.52689/1.00 so that [A-] = 0.52689 and [HA] = 1.00
Now calculate the decimal part of each component above.
A- = 0.52689/(0.52689 + 1.00) = 0.34507
and HA = 1.00/(0.52689 + 1.00) = 0.654926
Now this is where we are stuck. We found out the fraction of each component present in the buffer, but we cannot proceed further unless we know the molarity of the buffer. Since you are asked to write out a procedure, let us assume you want to make a 0.1 M buffer solution.
Then we calculate molarity of each component in the buffer by multiplying the fraction of each component by the total molarity of the buffer.
MA- = (0.1 M)*(0.34507) = 0.034507 M
and MHA = (0.1 M)*(0.654296) = 0.0654296 M
Next calculate the moles of each buffer component by multiplying by the total volume of the buffer. The volume of the buffer is 250.00 mL = (250.00 mL)*(1.00 L/1000.00 mL) = 0.25 L
mA- = (0.034507 mol/L)*(0.25 L) = 8.62675*10-3 moles
mHA = (0.0654296 mol/L)*(0.25 L) = 0.0163574 moles
Now, we have calculated the moles of each component that goes into the buffer. We need to calculate the weight or volume of each component to be added. Sodium bicarbonate is a white powder having molar mass 84 gm.
Therefore, the weight of sodium bicarbonate to be taken = (0.0163574 mole)*(84 gm/1 mole) = 1.374 gm
Let us say we have a stock solution of 1 M carbonic acid. Then volume of carbonic acid required = (8.62675*10-3 mole)/(1 mol/L) = 8.62675*10-3 L = (8.62675*10-3 L)*(1000 mL/1 L) = 8.62675 mL 8.63 mL
Therefore, to prepare 250.00 mL of 0.1 M bicarbonate buffer, we must accurately weight out 1.374 gm of sodium bicarbonate and take 8.63 mL of stock 1 M carbonic acid solution. The two are then added to a 250 mL volumetric flask and the volume is made upto the mark with distilled water.