A 44.7 mL sample of a 0.280 M solution of NaCN is titrated by
0.220 M...
90.2K
Verified Solution
Link Copied!
Question
Chemistry
A 44.7 mL sample of a 0.280 M solution of NaCN is titrated by0.220 M HCl. Kb for CN- is 2.0×10-5. Calculate the pH of thesolution:
(a) prior to the start of the titration
(b) after the addition of 28.4 mL of 0.220 M HCl
(c) at the equivalence point
(d) after the addition of 83.6 mLof 0.220 M HCl.
PLEASE SHOW ALL WORK
Answer & Explanation
Solved by verified expert
4.0 Ratings (609 Votes)
a prior to the start of the titration molarity of NaCN 0280 M CN H2O HCN OH 0280 0 0 initial 0280 x x x Kb x2 0280 x 20 x 105 x2 0280 x x 236 x 103 x OH236 x 103 pOH log OH pOH log 236
See Answer
Get Answers to Unlimited Questions
Join us to gain access to millions of questions and expert answers. Enjoy exclusive benefits tailored just for you!
Membership Benefits:
Unlimited Question Access with detailed Answers
Zin AI - 3 Million Words
10 Dall-E 3 Images
20 Plot Generations
Conversation with Dialogue Memory
No Ads, Ever!
Access to Our Best AI Platform: Zin AI - Your personal assistant for all your inquiries!
