Acids Acid Formula Ionization Constant K a Acetic CH 3 COOH 1.8×10 –5 Arsenic H

Question

Acids Acid Formula Ionization Constant Ka Acetic CH3COOH 1.8×10–5 Arsenic H3AsO4 5.5×10–3 Benzoic C6H5COOH 6.3×10–5 Boric H3BO3 5.4×10–10 Butanoic C3H7COOH 1.5×10–5 Carbonic H2CO3 Ka1 = 4.5×10–7
Ka2 = 4.7×10–11 Chloroacetic ClCH2COOH 1.4×10–3 Chlorous HClO2 1.1×10–2 Chromic H2CrO4 Ka1 = 1.8×10–1
Ka2 = 3.2×10–7 Cyanic HCNO 3.5×10–4 Formic HCOOH 1.8×10–4 Hydrazoic HN3 2.5×10–5 Hydrocyanic HCN 6.2×10–10 Hydrofluoric HF 6.3×10–4 Hydrogen peroxide H2O2 2.4×10–12 Hydrosulfuric H2S Ka1 = 8.9×10–8
Ka2 = 1.0×10–19 Hypobromous HBrO 2.8×10–9 Hypochlorous HClO 4.0×10–8 Hypoiodous HIO 3.2×10–11 Iodic HIO3 1.7×10–1 Nitrous HNO2 5.6×10–4 Oxalic C2H2O4 Ka1 = 5.6×10–2
Ka2 = 1.5×10–4 Paraperiodic H5IO6 2.8×10–2 Pentanoic C4H9COOH 1.5×10–5 Periodic HIO4 7.3×10–2 Phenol HC6H5O 1.3×10–10 Phosphoric H3PO4 Ka1 = 6.9×10–3
Ka2 = 6.2×10–8
Ka3 = 4.8×10–13 Phosphorous H3PO3 Ka1 = 5.0×10–2
Ka2 = 2.0×10–7 Propanoic C2H5COOH 1.3×10–5 Sulfuric H2SO4 Ka1 = very large
Ka2 = 1.2×10–2 Sulfurous H2SO3 Ka1 = 1.4×10–2
Ka2 = 6.3×10–8 Trichloroacetic Cl3CCOOH 2.2×10–1 Question 4 of 5 Sapling Learning Using this table of ionization constants, calculate the overall equilibrium constant, Koveral, for the following reaction. HBrolaq)+ No, (aq) Bro (aq) ,(aa) HNO, Number K.weran about us l careers l privacy policy l terms of use contact us l help Map

Explanation / Answer

K1:

HBrO <--< H+ + BrO-

Ka1 = [H+][BrO-]/[HBrO]

K2:

HNO2 <--> H+ + NO2-

Ka = [H+][NO2-]/[HNO2]

so

Ktotal = Ka1 * 1/Ka2

Ktotal = (2.8*10^-9) * 1 /(5.6*10^-4)

Ktotal = 0.000005

which is not favoured, since HNO2 is stronger acid, so expect NO2- in solution, not HNO2

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