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A 44.7 mL sample of a 0.280 M solution of NaCN is titrated by 0.220 M...

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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
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