Asked by Angel
A volume of 500.0 mL of 0.170 M NaOH is added to 595 mL of 0.200 M weak acid (𝐾a=6.74×10−5). What is the pH of the resulting buffer?
HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
Answers
Answered by
DrBob222
mols HA = M x L = 0.200 x 0.595 = 0.119
mols NaOH = 0.170 x 0.500 = 0.085
So you form 0.085 mols NaA and you have remaining 0.119 - 0.085 = 0.034 mols HA.
(acid) = mols/L = 0.034/(0.500 + 0.595) = 0.034/1.095 = ?
(base) = the salt = 0.085/1.095 = ?
pH = pKa + log [(base)/acid)]
Substitute the data and calculate the pH of the solution. Post your work if you get stuck.
mols NaOH = 0.170 x 0.500 = 0.085
So you form 0.085 mols NaA and you have remaining 0.119 - 0.085 = 0.034 mols HA.
(acid) = mols/L = 0.034/(0.500 + 0.595) = 0.034/1.095 = ?
(base) = the salt = 0.085/1.095 = ?
pH = pKa + log [(base)/acid)]
Substitute the data and calculate the pH of the solution. Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.