1) Rank the following titrations in order of increasing pH at the halfway point to equivalence...

1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH).

1 2 3 4 5  200.0 mL of 0.100 M H₂NNH₂ (K_b = 3.0 x 10^-6) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M HC₃H₅O₂ (K_a = 1.3 x 10^-5) by 0.100 M NaOH

1 2 3 4 5  100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH

1 2 3 4 5  200.0 mL of 0.100 M (C₂H₅)₂NH (K_b = 1.3 x 10^-3) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M HI by 0.100 M NaOH

2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH).

1 2 3 4 5  100.0 mL of 0.100 M HC₃H₅O₂ (K_a = 1.3 x 10^-5) by 0.100 M NaOH

1 2 3 4 5  200.0 mL of 0.100 M (C₂H₅)₂NH (K_b = 1.3 x 10^-3) by 0.100 M HCl

1 2 3 4 5  200.0 mL of 0.100 M H₂NNH₂ (K_b = 3.0 x 10^-6) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH

1 2 3 4 5  100.0 mL of 0.100 M KOH by 0.100 M HCl

Solutions

Expert Solution

1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH).

Solution :- larger the kb means higher the pH

So the smallest kb will have lower pH so the order is as follows

  4= 200.0 mL of 0.100 M H₂NNH₂ (K_b = 3.0 x 10^-6) by 0.100 M HCl

  5= 100.0 mL of 0.100 M HC₃H₅O₂ (K_a = 1.3 x 10^-5) by 0.100 M NaOH

3=  100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH
2=   200.0 mL of 0.100 M (C₂H₅)₂NH (K_b = 1.3 x 10^-3) by 0.100 M HCl

1 =   100.0 mL of 0.100 M HI by 0.100 M NaOH

2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH).

5=  100.0 mL of 0.100 M HC₃H₅O₂ (K_a = 1.3 x 10^-5) by 0.100 M NaOH

2=  200.0 mL of 0.100 M (C₂H₅)₂NH (K_b = 1.3 x 10^-3) by 0.100 M HCl

1=  200.0 mL of 0.100 M H₂NNH₂ (K_b = 3.0 x 10^-6) by 0.100 M HCl

4=  100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH

3 =  100.0 mL of 0.100 M KOH by 0.100 M HCl

queen_honey_blossom answered 1 year ago

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