1
answer
0
watching
744
views
17 Nov 2019
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 KOH by 0.100 M HCl 200.0 mL of 0.100 M HC_2H_3O_2 (K_a = 1.8 times 10^-5) by 0.100 M NaOH 100.0 mL of 0.100 M HCl by 0.100 M NaOH 200.0 mL of 0.100 M H_2NNH_2 (K_b = 3.0 times 10^-6) by 0.100 M HCl 100.0 mL of 0.100 M C_2H_5NH_2 (K_b = 5.6 times 10^-4) by 0.100 M HCl 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 HNO_2 (K_a = 4.0 times 10^-4) by 0.100 M NaOH 100.0 mL of 0.100 M C_2H_5NH_2 (K_b = 5.6 times 10^-4) by 0.100 M HCl 200.0 mL of 0.100 M HC_2H_3O_2 (K_a = 1.8 times 10^-5) by 0.100 M NaOH 200.0 mL of 0.100 M H_2NNH_2 (K_b = 3.0 times 10^-6) by 0.100 M HCl 100.0 mL of 0.100 M KOH by 0.100 M HCl
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 KOH by 0.100 M HCl 200.0 mL of 0.100 M HC_2H_3O_2 (K_a = 1.8 times 10^-5) by 0.100 M NaOH 100.0 mL of 0.100 M HCl by 0.100 M NaOH 200.0 mL of 0.100 M H_2NNH_2 (K_b = 3.0 times 10^-6) by 0.100 M HCl 100.0 mL of 0.100 M C_2H_5NH_2 (K_b = 5.6 times 10^-4) by 0.100 M HCl 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 HNO_2 (K_a = 4.0 times 10^-4) by 0.100 M NaOH 100.0 mL of 0.100 M C_2H_5NH_2 (K_b = 5.6 times 10^-4) by 0.100 M HCl 200.0 mL of 0.100 M HC_2H_3O_2 (K_a = 1.8 times 10^-5) by 0.100 M NaOH 200.0 mL of 0.100 M H_2NNH_2 (K_b = 3.0 times 10^-6) by 0.100 M HCl 100.0 mL of 0.100 M KOH by 0.100 M HCl
Jamar FerryLv2
6 Mar 2019