1. Student Q prepared 50.00 mL of a buffer solution using 0.50 moles of HA and 0
ID: 830831 • Letter: 1
Question
1. Student Q prepared 50.00 mL of a buffer solution using 0.50 moles of HA and 0.50 moles of A- while Student S prepared 50.00 mL of a buffer solution using 0.25 moles of HA and 0.25 moles of A-.
a. Do the two buffer solutions have the same or different pH? Explain.
b. If 1.00 mL of 0.010M NaOH were added to the two buffer solutions, would the pH of the two solutions increase or decrease? Explain.
c. Which solution (Buffer Q or Buffer S) would show a smaller change in the pH for question (b)? Explain.
Please give answers to all parts and EXPLAIN and I will rate :) ... thank you
Explanation / Answer
(a) The two solutions will have the same pH since the relative moles of A- : HA are the same.
Initial pH buffer Q = pKa + log([A-]/[HA])
= pKa + log(moles of A-/moles of HA)
= pKa + log(0.50/0.50) = pKa
Initial pH buffer S = pKa + log([A-]/[HA])
= pKa + log(moles of A-/moles of HA)
= pKa + log(0.25/0.25) = pKa
(b) The pH of the two solutions will increase as NaOH is a base and will decrease the net concentration of H+ in the solution
Moles of NaOH added = 1.00/1000 x 0.010 = 0.00001 mol
HA + NaOH => Na+ + A- + H2O
For solution Q:
Moles of HA = 0.50 - 0.00001 = 0.49999 mol
Moles of A- = 0.50 + 0.00001 = 0.50001 mol
Final pH buffer Q = pKa + log([A-]/[HA])
= pKa + log(moles of A-/moles of HA)
= pKa + log(0.50001/0.49999) = pKa + 1.737 x 10^(-5)
For solution S:
Moles of HA = 0.25 - 0.00001 = 0.24999 mol
Moles of A- = 0.25 + 0.00001 = 0.25001 mol
Final pH buffer S = pKa + log([A-]/[HA])
= pKa + log(moles of A-/moles of HA)
= pKa + log(0.25001/0.24999) = pKa + 3.474 x 10^(-5)
Since final pH > initial pH => pH increases
(c) From calculation above:
Solution Q shows a smaller pH change (as it as a higher buffering capacity)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.