bsuffer wish a specified plt and opacity luffers are unique solutions in which a
ID: 964125 • Letter: B
Question
bsuffer wish a specified plt and opacity luffers are unique solutions in which a weak acid is in equilibrium with its conjugate base addition of small amounts of hase will not disturb the pli of the solution much. The weak acid and base f until the equivalence point is reached for each and for determination lenderson-FHasselbalch equation t to observe changes in buffer pacity and the balf-way point of the weak acid and base in the titrations PRE-LAB ASSIGNMENT ) Read this lab handout in its entirety 2) Wear appropriate shoes to lab 3) Suppose you would like to make 100.00 mL of a pH acetic acid/acetate the acid concentration is 903 M. What concentration 4) How many ml of 1.00 M acetic acid are needed for the buffer? pH acetic acid/acetate buffer, in which M. What concentration of sodium acetate is needed in this buffer? 4 buffer? 5) How many grams of sodium acetate are needed for the 6) Write the reaction of the buffer with HCI solution How many mL of 0.100 M HCI can react with the buffer before it is exhausted? Write the reaction of the buffer with NaOH solution. How many mL of 0.100 M NaOH can react with the buffer before it is exhausted? PROCEDURE SAFETY: Avoid skin contact with all solutions. Acids and bases are skin irritants. Clean up Ils Buffer . Add 50 mL s completely immersed. Allow the pH to stabilize and record the value . Add I drop of 0.100 M HCI to the beaker and mix the solution. Allow the pH to stabilizse and record the value To a clean beaker, add 50 mL of water and the pH probe. Allow the pH to stabilize and record the value. .Add I drop of 0.100 M NaOH to the beaker and mix the solution. Allow the pH to stabilize Add I drop of 0.100 M NaOH to the beaker and mix the solution. Allow the pH to and record the value 10-+ do2s- OSExplanation / Answer
I can't read very well the concentration in part 3, I think it said 0.01 M right? if it's not, just change the value in the calculations:
3. pH = pKa + log[A]/[HA] pKA of acetic acid = 4.74
4.76 = 4.74 + log[A]/0.010
0.02 = logA/0.01
100.02 = A/0.01
1.0471 = A/0.01
[A] = 0.01047 M
4. I'll leave this question to you, just use dillution factor.
5. The grams of the acetate, we already know the concentration so the moles:
moles A = 0.01047 * 0.1 = 0.001047 moles
mass A = 0.001047 * 82.0343 = 0.0859 g
6. H+ + CH3COO- -------> CH3COOH + H2O
Adding HCl means that the solution will increase the acid concentration so:
Va = 0.001 moles of HCl * 1 L / 0.1 M = 0.01 L or simply 10 mL
In the base occurs the same but with the moles of the acetate. The reaction:
CH3COOH + OH- --------> CH3COO- + H2O
Vb = 0.001047 * 1 L / 0.1 = 0.01047 L or 10.47 mL
Hope this helps
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