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

ID: 1009121 • Letter: R

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 HCI 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 HNO_2 (K_a = 4.0 times 10^14) by 0.100 M NaOH 100.0 mL of 0.100 M C_2H_5NH_2 (K_b = 5.6 times 10^14) by 0.100 M HCI 100.0 mL of 0.100 M NH_3 (K_b = 1.8 times 10^-5) 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 (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M C_2H_5NH_2 (K_b = 5.6 times 10^14) by 0.100 M HCI 100.0 mL of 0.100 M HN0_2 (K_a = 4.0 times 10^14) by 0.100 M NaOH 100.0 mL of 0.100 M HC_3H_5O_2 (K_a = 1.3 times 10^15) 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 NH_3 (K_b = 1.8 times 10^15) by 0.100 M HCI

Explanation / Answer

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HCl and NaOH will react to form neutral solution
pH = 7

HCl and KOH will react to form neutral solution
pH = 7

Between HNO2 and NaOH, NaOH is major species
NaOH is base anmd hence pH will be more than 7

Between C2H5NH2 and HCl, HCl is major species
HCl is acis so pH will be less than 7

Between NH3 and HCl, HCl is major species
HCl is acis so pH will be less than 7

NH3's conjugate acis will have greater Ka and hence more acidic. So pH will be least here

order will be:
3
3
5
2
1