you will be working originally with 75.0 mL of a 0.20 M solution of HC4H7O2 (Ka= 1.5 x 10-5 ) 1. Determine the pH of your solution (I got pH= 2.7 with significant figures taken into account, is that right?) 2. Add 85.0 mL of 0.15 M KC4H7O2 to your solution. Find the pH. 3. Add 9.0 mL of 0.100 M HBr to the solution from task #2. Find the pH. 4. In a titration, 5.0 mL of 0.75 M KOH is added to your original solution. Find the pH. 5. What volume of 0.75 M KOH must be added to the original solution to produce a solution which would act as a buffer more than any other time in the titration? What would the pH of the solution be at that point? 6. What volume of 0.75 M KOH must be added to the original solution to reach the equivalence point in the titration? What would the pH of the solution be at that point? 7. If a volume of 25.0 mL of 0.75 M KOH is added to the original solution in a titration, what would the pH of the solution be at that point? 8.If a weak base was used in place of the weak acid in the titration and a strong acid was used in place of the strong base referenced in tasks 4-7 (same volumes, same concentrations) name TWO noteworthy and appropriate similarites in the process in the first scenario [for which you did the work above] compared to this new scenario and TWO noteworthy and appropriate differences as well. PLEASE ANSWER THOROUGHLY AND TRY TO SHOW WORK. THANK YOU!!
1) HC4H7O2 ---------> C4H7O2- + H+ , , [H+] =[C4H7O2-] = x/(0.075),
[HC4H7O2] = (0.2x0.075-x)/0.075 = (0.015-x)/0.075
Ka = [H+][C4H7O2-]/[HC4H7O2] = x^2/(0.015-x)(0.075) = 1.5 x10^ -5
x = 0.0004 = H+ moles , [H+] = (0.0004/0.075) = 0.00533
pH = -log( 0.00533) = 2.27
2) total vol = 75+85 = 160 ml = 0.16, pka = -log(1.5 x10^ -5) = 4.824
pH = pka + log[KC4H7O2]/[HC4H7O2] ,
[KC4H7O2] = (0.085x0.15)/(0.16 = 0.08, [HC4H7O2] = (0.075x0.2)/0.16 =0.09375
pH = 4.824 + log( 0.08/0.09375)
= 4.755,
3) 9 ml of 0.1 M HBr added , [KC4H7O2] = [( 0.08x0.16-(0.1x0.009)]/(0.169) = 0.0704,
[HC4H7O2] = (0.09375 x 0.16 +(0.1x0.009)]/0.169 = 0.094,
pH = 4.824 + log( 0.0704/0.094)
= 4.698,
4) 5 ml of 0.75 M KOH added
[KC4H7O2] = ( 0.75 x 0.005)/(0.08) = 0.046875, [HC4H7O2] = [( 0.075x0.2-(0.75x0.005)]/0.08 =0.14,
pH = 4.824 + log( 0.046875/0.14)
= 4.345,
5) buffer has max capacity when pH = pka ,
hence when pH = pka we get [KC4H7O2] =[HC4H7O2]
0.75 xV = [( 0.075x0.2-(0.75xV)]
V =0.01 liter = 10 ml
6) PH at this poin = pka = 4.824
7) excess KOH = ( 0.75x0.025-(0.075x0.2) /0.15 = 0.025,
pOH = -log( 0.025) = 1.6,
pH = 12.4
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