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  3. consider the titration of 80.0 ml of 0.0200 m c5h5

Consider the titration of 80.0 mL of 0.0200 M C5H5N (a weak base; Kb = 1.70e-09)...

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Chemistry

Consider the titration of 80.0 mL of 0.0200 M C5H5N (a weakbase; Kb = 1.70e-09) with 0.100 M HNO3.

Calculate the pH after the following volumes of titrant havebeen added:

(a) 0.0 mL pH =

(b) 4.0 mL pH =

(c) 8.0 mL pH =

(d) 12.0 mL pH =

(e) 16.0 mL pH =

(f) 25.6 mL pH =

PART TWO:

A buffer solution contains 0.14 mol of ascorbic acid(HC6H7O6) and 0.84 mol of sodiumascorbate (NaC6H7O6) in 2.80L.
The Ka of ascorbic acid(HC6H7O6) is Ka =8e-05.



(a) What is the pH of this buffer?

pH =  


(b) What is the pH of the buffer after the addition of 0.07 mol ofNaOH? (assume no volume change)

pH =  


(c) What is the pH of the original buffer after the addition of0.21 mol of HI? (assume no volume change)

pH =  

Answer & Explanation Solved by verified expert
3.7 Ratings (291 Votes)
I will answer Part 1 I highly reccomend you to post your second question in another post to get answered First lets calculate the equivalence point volume Va 002 80 01 16 mL This means that in 16 mL we reached the equivalence point Beyond this point we have excess of acid a At 0 mL no acid is added only base    See Answer
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