Rank The following titrations in order of increasing pH at The halfway point to

Question

Rank The following titrations in order of increasing pH at The halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HOCI (Ka = 3.5 x 10-8) by 0.100 M NaOH 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCI 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 x 10-4) by 0.100 M HCI 100.0 mL of 0.100 M HCI by 0.100 M NaOH Consider The major species present at The halfway point. From this, you should be able to deduce The order of The pH with only minimal calculations Rank The following titrations in order of increasing pH at The equivalence point of The titration (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HOCI (Ka = 3.5 x 10-8) by 0.100 M NaOH 100.0 mL of 0.100 M KOH by 0.100 M HCI 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 x 10-4) by 0.100 M HCI 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCI 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH

Explanation / Answer

at halfwaypoint to equivalence point

pH = pKa or pKb of weakacid or base.

i. pka = log ka = -log(3.5*10^(-8)) = 7.45

pH = 7.45

ii. pH = -log(3.0*10^(-6)) = 5.523

iii. pH = -log(7.2*10^(-4)) = 3.14

iv . pH = -log(5.6*10^(-4)) = 3.25

v. concentration of NaOH = 100/1000*0.1 = 0.01 M

poH = -log[H+] = -log(0.01)   = 2

pH = 12

ranking   1.iii, 2.iv, 3.ii, 4.i , 5.v

2) i. at equivalence point pH = 7+1/2(pka+logc)

     = 7+1/2(7.45+log0.01)
  
= 9.75

ii.KOH +HCl --->KCl+ H2o

pH = 7

iii. pH = = 7+1/2(3.25+log0.01) = 7.625

iv . pH = = 7+1/2(5.52+log0.02) = 8.91

v. ph = 7+1/2(3.14+log0.01) =7.57

ranking 1.ii , 2. v , 3. iii, 4.iv, 5.i

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