20.01 6) Compare the change in pll that occurs when added to 100 ml of a benzoic

Question

20.01 6) Compare the change in pll that occurs when added to 100 ml of a benzoic results. of s benzoiec acid buffer solutionhat is 0.15 M in both C,H,co,H and C,H,CO Na. Explain the 7) Calculate the pH during the titration of 35.00 ml. of 0.150 M HCOOH with 0.250 M NaOH after the addition of 0.00 ml, 10.50 ml, 21.00 ml, and 30.00 ml of the base. Sketch the graph for this process and label the curve with as much detail as possible. 8) Calculate the pH during the titration of 50.0 ml. of a 0.075 M HNO, solution with a 0.1 M KOH solution, after the addition of 0.00 ml, 10.50 ml, 37.5 ml, and 50.0 ml of the base. Sketch the graph for this process and label the curve with as much detail as possible. 9) Explain the difference in pH's observed at the equilivelence points for the titrations in questions 7 and 8 esung.edu/K/konzelman./CHEM%201212/Exam%2...Oand%20Bases/wksht%20Quantitative%20acid%20 base%20.doc

Explanation / Answer

Q6

pH in pure waer = 7

pH after

mmol of NaOH= MV = 0.1*10 = 1

Vtotal = 100+10 = 110

[NaOH new] = mmol/V = 1/110 = 0.00909

[OH-] = 0.00909

pOH = -log(0.00909 = 2.0414

pH = 14-2.0414 = 11.9586

pH change = 11.9586-7

dpH = 4.9586

B)

siilar to a buffer:

mmol of base = 1

mmol of buffer:

mmol of HA = 0.15*100 = 15

mmol of A- = 0.15*100 = 15

after reaction with base

mmol of HA = 15-1 = 14

mmol of A- = 15+1 = 16

pH = pKa + log(A-/HA)

pKa for benzoic acid = 4.20

pH = 4.2 + log(16/14) = 4.2579

pH initial when 0.15 M each:

pH = 4.2 + log(0.15/0.15) = 4.2

then

dpH = 4.2579 - 4.20 = 0.0579

the pH will not change since this is abuffer, will form equilibiria, compared with the water alone

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