You want to make a Tris buffer stock solution for future use in your lab. You cu
ID: 713712 • Letter: Y
Question
You want to make a Tris buffer stock solution for future use in your lab. You currently have both Tris acid (MW: 157.6) and Tris base (MW: 121.14) available to use. The pka of this buffer system is 8.3. 1. If you want to have a final volume of 750mL for a 0.5M solution. How much mass would you need to make each of these solutions? (Assume you are making one solution of Tris acid and one of Tris base). (10 points) a. b. From these stock solutions you want to make a working solution of your Tris buffer with a final volume of 50mL with a concentration of 20mM and a pH of 8.6. How much of each solution do you need to make this buffer? (10 points) Your friend is running a similar experiment and wants to borrow some of your buffer instead of making their own. If they want the pH to be 5.5, would you agree they should use your buffer or suggest they make their own. Why? (5 points) c. d. When you performed the Acids and Bases lab, what was the pH of your buffer from part 2? What page of your notebook is this on? (5 points)Explanation / Answer
a) V=750ml=0.750L
Ct=final concentration=0.5M=0.5mol/L
mol of tris acid required=0.750L*0.5mol/L=0.375mol
mass of tris acid required=mol*molar mass of tris acid=0.375mol/L*(157.6g/mol)=59.1g g
mass of tris base required=mol*molar mass of tris base=0.375mol/L*(121.14g/mol)=45.427g g
b)buffer with pH=8.6
pka=8.3
Using Henderson-hasselbach equation,
pH=pka+log [base]/[acid]
8.6=8.3+log [base]/[acid]
[base]/[acid]=1.995
So,the ratio of mol of acid and base should be 1.995
total mol to be prepared=50ml*20mM=0.050L*20*10^-3 mol/L=0.001mol
mol of tris acid required=(1/1+1.995)*0.001mol=0.000334mol
mol of tris base required=(1.995/1+1.995)*0.001mol=6.661*10^-4mol
volume of stock tris acid required for the buffer=mol/molarity=0.000334mol/0.5mol/L=0.000668L=0.668ml
volume of stock tris base required for the buffer=mol/molarity=6.661*10^-4mol/0.5mol/L=0.00133L=1.33ml
c)required pH=5.5,
For lowering pH concentrated acid ,like HCl ,needs to be added.that neutralizes tris base to give more tris acid,decreasing pH
Calculation of amount of HCl to be added:
pH=pka+log [base]/[acid]
or,5.5=8.3+log [base]/[acid]
[base]/[acid]=0.00158
total mol of bufer=50ml*20mM=0.050L*20*10^-3 mol/L=0.001mol
[tris base]=(0.00158/1.00158)*0.001mol=1.577*10^-6 mol
[tris acid]=(1/1.00158)*0.001mol=9.984*10^-4 mol
6.661*10^-4mol of tris base must be neutralized to 1.577*10^-6 mol by adding strong acid
mol of strong acid to be added=(6.661*10^-4)-(1.577*10^-6)=6.645*10^-4 mol
This ,also adds 6.645*10^-4 mol to the tris acid,now tris acid is =0.000334mol+6.645*10^-4 mol=9.985*10^-4 mol
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