Consider 500 mL of a buffer 0.12 M HoCl and 0.080 M KOCl. Do you predict the pH
ID: 487042 • Letter: C
Question
Consider 500 mL of a buffer 0.12 M HoCl and 0.080 M KOCl. Do you predict the pH of the buffer to above or below the pK _a.? Above the pK _a Below the pK _a EXPLAIN your choice: Given that HOCl has a K _a = 3.5 times 10^-8 calculate the pH of this buffer? Does your calculated value match your prediction? Now, let's add 2.5 mL of 10 M KOH to this buffer Do you predict the pH of the buffer to increase, decrease or stay the same when you add KOH? Increase Decrease Stay the same EXPLAIN your choice: Finally, let's calculate the pH of the buffer after the addition of the strong base. To approach this problem follow the following 4 step procedure: determine the number of mole of HOCl, OH^- and OCl^- that you begin with (Remember that you must work in moles, not molarity, for a limiting reagent problem) determine the limiting reagent and use the ICE table to calculate the number o moles of HOCl, OH^- and OCl^- that are present in the solution after the reaction calculate the concentrations of these species in units of molarity after the reaction. Use the species present after the reaction to calculate the new pH.Explanation / Answer
2.
a)HOCL = H+ + KOCL
Since the acid form HOCl (0.12 M) concentration is higher than the KOCl (0.08 M), the pH is less than its pKa.
At pKa, we will have equal conentration of acid and basic form.
b) pH = pKa + log ([OCl-]/[HOCl]
pH = -log (3.5 x 10^-8) + log (0.08/0.12) = 7.28
pH (7.28) is less the pKa (7.45); our prediction is matching with our calculations.
c) Here, we are adding considerable amounts of KOH to buffer solution, 0.025 moles of KOH. This will dissociate
HOCl to ClO- and leading to more KOCl concentration. Therefore, the pH of the solution will increase.
d) You are adding 0.0025 L X 10 mol/L = 0.025 moles KOH to the buffer.
It will convert 0.025 moles HOCl into 0.025 moles more of ClO-.
Initial moles HOCl = 0.5 L X 0.12 mol/L = 0.06 mol HClO
Initial moles ClO- = 0.5 L X 0.08 mol/L = 0.04 mol ClO-
After the addition of KOH, moles HOCl = 0.035 and moles ClO- = 0.065
Final molarity HOCl = 0.07 M, final molarity ClO- = 0.13 M
Then, pH = pKa + log (0.13/0.07) = 7.718
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