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

 

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1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 4 200.0 ml of 0.100 M (C,Hs), NH (K, - 1.3 x 10) by 0.100 M HCI 1 100.0 ml. of 0.100 M KOH by 0.100 M HCI 2 100.0 ml. of 0.100 M HC Hs02 (K, = 1.3 x 10) by 0.100 M NaOH 5 100.0 mL of 0.100 M HI by 0.100 M NAOH 3 200.0 ml. of 0.100 M H,NNH, (Ke - 3.0 x 10) 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 Submit Answer Incorrect. Tries 2/45 Previous Tries 2) Rank the folowing 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 KOH by 0.100 M HCI 200.0 ml of 0.100 M H,NNH, (Ke- 3.0 x 106) by 0.100 M HCI 100.0 ml of 0.100 M HF (K, 7.2 x 10) by 0.100 M NaOH v 100.0 ml of 0.100 M HC3HsO, (K, = 1.3 x 10) by 0.100 M NaOH 200.0 ml. of 0.100 M (CH),NH (K, = 1.3 x 10) by 0.100 M HCI 1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 4 200.0 ml of 0.100 M (C,Hs), NH (K, - 1.3 x 10) by 0.100 M HCI 1 100.0 ml. of 0.100 M KOH by 0.100 M HCI 2 100.0 ml. of 0.100 M HC Hs02 (K, = 1.3 x 10) by 0.100 M NaOH 5 100.0 mL of 0.100 M HI by 0.100 M NAOH 3 200.0 ml. of 0.100 M H,NNH, (Ke - 3.0 x 10) 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 Submit Answer Incorrect. Tries 2/45 Previous Tries 2) Rank the folowing 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 KOH by 0.100 M HCI 200.0 ml of 0.100 M H,NNH, (Ke- 3.0 x 106) by 0.100 M HCI 100.0 ml of 0.100 M HF (K, 7.2 x 10) by 0.100 M NaOH v 100.0 ml of 0.100 M HC3HsO, (K, = 1.3 x 10) by 0.100 M NaOH 200.0 ml. of 0.100 M (CH),NH (K, = 1.3 x 10) by 0.100 M HCI

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