Please help! Rank the following titrations in order of increasing pH at the half

Question

Please help! Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 5 100.0 of 0.100 M HCI by 0.100 M NaOH 4 100.0 ml of 0.100 M HOCI (Ka = 3.5 times 108) by 0.100 M NaOH 2 100 0 mL of 0.100 M C2H5NH2 (Ka = 5.6 times 10-4) by 0.100 M HCI 3 100.0 mL of 0.100 M KOH by 0.100 M HCI 1 200 0 mL of 0.100 M H2NNH2 (Ka - 3.0 times 10-6) by 0.100 M HCI 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 (l = lowest pH and 5 - highest pH). 5 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 times 10 6) by 0.100 M HCI 4 B 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 times 10'4) by 0.100 M HCI 2 100.0 mL of 0.100 M KOH by 0.100 M HCI 3 100.0 mL of 0.100 M HF (Ka = 7.2 times 104) by 0.100 M NaOH 1 100.0 mL of 0.100 M HOCI (Ka = 3.5 times 10 8) by 0.100 M NaOH Consider the major species present at the equivalence point. From this, you should be able to deduce the order of the pH with only minimal calculations

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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