Build a balanced picture on the dissociation of the following the inside liquids and select the conjugate acid-base pairs

  • HNO3
  • Ba(OH)2
  • HlPOcuatro
  • CH3COOH

At a particular temperature, the latest K

  1. NH4
  2. H2SO4
  3. CH3COOH.
  • CaO – Lewis foot – All of the gold and silver oxides is actually Lewis angles
  • CO2 – Lewis acid – CO2 contains a polar double bond.

Question 4. H3BO3 accepts hydroxide ion from water as shown below Answer: H3BO3 (aq) + H2O(l) = B(OH)4 – + H + Predict the nature of H3BO3 using Lewis concept. Boric acid is also called as hydrogen borate or orthoboric acid. It is a weak mono basic Lewis acid of boron and it is written as B(OH)3. It accepts hydroxyl (OH – ) ion from water. It does not dissociate to give hydronium (H3O + ) ion rather forms metaborate ion and this ions in turn give H3O ion. B(OH)3 + H2O [B(OH)4] – + H3O + Hence it is considered as weak acid.

Question 5. w of a neutral solution was equal to 4 x ten -14 . Calculate the concentration of [H3O + ] and [OH – ]. Answer: Given solution is neutral [H3O + ] = [OH – ] Let [H3O + ] = x ; then [OH – ] = x Kw = [H3O + ] [OH – ] 4 x 10 -14 = x . x x 2 = 4 x 10 -14

What’s the improvement in this new pH once adding 0

In this case the concentration of H2SO4 is very low and hence [H3O] from water cannot be neglected [H3O + ] = 2 x 10 -8 (from H2SO4) + 10 -7 (from water) = 10 -8 (2+ 10) = 12 x 10 -8 = 1.2 x 10 -7 pH = – log10[H3O + ] = – log10( 1.2 x 10 -7 ) = 7 – log101.2 = 7 – 0.0791 = 6.9209

2. pH of the solution = 5.4 [H3O + ] = antilog of (- pH) = antilog of (- 5.4) = antilog of (-6 + 0.6) = \(\overline <6>.6\) = 3.981 x 10 -6 i.e., 3.98 x 10 -6 mol dm -3

step 3. No. of moles regarding HCl = 0.2 x fifty x 10 -3 = ten x 10 -step three No. off moles out-of NaOH =0.step one x 50 x 10 -step three = 5 x 10 -3 Zero. from moles of HCl immediately after mix = 10 x ten -3 – 5 x 10 -step three = 5 x 10 -step three immediately after mixing full regularity = one hundred mL

[H3O + ] = escort service Mesquite 5 x 10 -2 M pH = – log ( 5 x 10 -2 ) = 2 – log 5 = 2 – 0.6990 = 1.30

Question 7. Kb for NH4OH is 1.8 x 10 -5 Calculate the percentage of ionisation of 0.06 M ammonium hydroxide solution. Answer:

2. Calculate the pH of a buffer solution consisting of 0.4M CH3COOH and 0.4 M CH3COONa . 01 mol of HCI to 500m1 of the above buffer solution.

Assume that the addition of HCI causes negligible change In the volume. Given: (K = 1.8 x 105). Answer: 1. Dissociation of buffer components NH4OH (aq) \(\rightleftharpoons\) NH4 + (aq) + OH – (aq) NH4CI > NH4 + + Cl – Addition of OH – The added H + ions are neutralized by NH4OH and there is no appreciable decrease in pH. NH4OH(aq) + H + \(\rightleftharpoons\) NH4 + (aq) + H2O (1) Addition of NH4 – (aq) + OH – (aq) > NH4OH (aq) The added OH ions react with NH4 to produce unionized NH4OH . Since NH4OH is a weak base, there is no appreciable increase in pH.

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