Generate a balanced formula on the dissociation of the adopting the for the liquids and you will choose this new conjugate acidic-legs sets

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

On a specific temperatures, the K

1. NH₄
2. H₂SO₄
3. CH₃COOH.

• CaO – Lewis foot – All of the gold and silver oxides is actually Lewis bases
• 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 Provo escort reviews 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 K_w = [H₃O + ] [OH – ] 4 x 10 -14 = x . x x 2 = 4 x 10 -14

What’s the improvement in the newest pH shortly after including 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 three. No. out-of moles from HCl = 0.dos x 50 x 10 -3 = 10 x 10 -3 Zero. off moles out-of NaOH =0.step one x 50 x 10 -3 = 5 x ten -step 3 Zero. out of moles of HCl shortly after mix = 10 x ten -step 3 – 5 x ten -3 = 5 x 10 -step 3 once fusion total volume = a hundred mL

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

Question 7. K_b 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.