General Chemistry Il 2. Consider 0.500 L of a buffer comprised of 0.12 M HOCI an
ID: 711612 • Letter: G
Question
General Chemistry Il 2. Consider 0.500 L of a buffer comprised of 0.12 M HOCI and 0.080 M KOCI. a) Do you predict the pH of the buffer to be above or below the pK? Below the pk EXPLAIN your choice b) Given that Hoa has a K,-3.5 x 10®. calculate the pH of this buffer? Does your calculated value match your prediction? c) Now, let's add 25 mL of 1.0M KOH to this buffer. Do you predict the pH of the buffer to increase, decrease or stay the same when you add KOH? Increase Stay the same EXPLAIN your choice: d) Finally, let's calculate the pH of the buffer after the addition of the strong base. HINT: This is a limiting reagent problem in which the strong base reacts with the weak acid component of the buffer. The reaction will go to completion, be consumed. The net chemical reaction you are interested in is: HOC1+ orocr + 110 so all the limiting reagent will Remember that for a limiting reagent problem you must work in moles, not molarity! Step 1: Determine the number of moles of HOCI, OH, and oCr that you begin with (don't forget to include the basic part of the buffer in your calculations!). Step 2: Determine the limiting reagent and then use the ICE (here Initial-Change-End) table to calculate the number of moles of HOCL, OH, and OCr that are present in the solution after the reaction. Step 3: Calculate the molarities of these species after the reaction. Step 4: Use the molarities of the species found in solution after the reaction to calculate the new pH. (Hint: Is the resulting solution still a buffer?)Explanation / Answer
a) Below the pKa
The concentration of acid [HOCl] >>the concentration of conjugate base,[OCl-],
pH is inversely proportional to [H3O+] or acid concentration ,so high value of [HOCl] will decrease pH
Mathematically,
ratio of [OCl-]/[HOCl] must be fractional giving log [OCl-]/[HOCl] as negative
Using Henderson-hasselbach equation,
pH=pka+log [OCl-]/[HOCl]=pka+(-ve value),pH<pka
b)ka=3.5*10^-8
pka=-log ka=-log (3.5*10^-8)=7.456
pH=pka+log(0.080/0.12)=7.456-0.176=7.28
pH=7.28 <pka
THe calculated value matches the prediction.
c)On addition of strong base ,HOCl will be neutralized to produce more OCl-
KOH+HOCl ---->KOCl +H2O
[OCl] will increase so basebconcentration will increase thus pH will also increase
d)mol of base -OCl present initially=0.5L*0.08mol/L=0.04 mol
mol of acid .HOCL present initially=0.5L*0.15mol/L=0.06 mol
mol of KOH added=mol of -OCl added=0.025L*1.0mol/L=0.025mol
mol of base present after acid neutrlization=0.04+0.025=0.065mol
mol of acid present after neutrlization txn =0.06-0.025=0.035 mol
[base]/[acid]=[OCl]/[HOCl]=0.065/0.035=1.857
pH=7.456+log 1.857=7.725
pH=7.7
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