(Chemical equilibria #6) I need the answer for part C. Part A: What volume of 10

Question

(Chemical equilibria #6) I need the answer for part C.
Part A: What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl?
Answer: 6.7 mL
Part B
Question: The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0250 mol of hydrogen ions without changing the volume. What is the pH of the final solution?
Answer: pH 7.05
Part C
Question: What additional volume of 10.0 M HCl would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B? In other words, how much more of this HCl solution is required to consume the remaining Tris in the buffer?

Express your answer in milliliters using two significant figures.
Answer in mL. Thank you

Explanation / Answer

part C , it is basic buffer ,

      pkb of Tris = 5.93

pOH = pkb + log(TrisHCl/tris)

buffer capacity pOH = (pkb+1) to (pkb-1)

no of mol of trisHCl taken = w/Mwt

               = 31.52/157.6

               = 0.2 mol

in half solution,

no of mol of trisHCl = 0.2/2 = 0.1 mol

No of mol of NaOH added = tris = 0.025 mol

by the addition of NaOH, pOH decreases.

maximum posibility = 5.93-1 = 4.94

so that,

4.93 = 5.93+log((0.1-0.025-x)/(0.025+x))

   x = 0.066

no of mol of NaOH must be added = 0.066 mol

vol.of NaOH must be added = n/M

                         = 0.066/10

                         = 0.0066 L

                         = 6.6 ml

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