Why don’t we check out the stabilities of one’s conjugate basics ClO

• Posted on 26/06/2022
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Such anions is stabilized so you’re able to higher extent, it offers smaller destination getting proton hence, usually work as poor feet. Consequently, the fresh new associated acid was strongest because the weak conjugate feet possess strong acidic and you will a strong conjugate feet have weak acid.

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This means ClO_cuatro – will have maximum stability and therefore will have a minimum attraction for W. Thus CIO₄ – will be weakest base and its conjugate acid HCIO₄ is the strongest acid.

When aqueous ammonia is added to CuSO₄ solution, the solution turns deep blue due to the formation of tetrammine copper (II) complex, [Cu(H₂O)_six] 2+ (aq) + 4NH_{step step three} (aq) \(\rightleftharpoons\) [Cu(NH₃)₄] 2+ (aq), among H₂O and NH₃ which is stronger Lewis base. Answer: Copper (II) sulphate solution, for example contains the blue hexaaqua copper (II) complex ion. In the first stage of the reaction, the ammonia acts as a Bronsted – Lowry base. With a small amount of ammonia solution, hydrogen ions are pulled off two water molecules in the hexaaqua ion. This produces a neutral complex, one carrying no charge.

If you remove two positively charged hydrogen ions from a 2+ ion, then obviously there isn’t going to be any charge left on the ion escort services in Murfreesboro. Because of the lack of charge, the neutral complex isn’t soluble in water and so you get a pale blue precipitate. [Cu(H₂O) 6 ] 2+ + 2NH₃ [Cu(H₂O)₄OH] + 2NH₄ +

This precipitate is often written as Cu(OH)₂ and called copper (II) hydroxide. The reaction is reversible because ammonia is only a weak base. That precipitate dissolves if you add an excess of ammonia solution, giving a deep blue solution. The ammonia uses its lone pair to form a coordinate covalent bond with the copper. It is acting as an electron pair donor – a Lewis base.

The fresh ammonia changes five of one’s h2o molecules inside the copper to offer tetramminc diaqua copper (II) ions

Question 6. The concentration of hydroxide ion in a water sample is found to be 2.5 x ten -6 M. Identify the nature of the solution. Answer: The concentration of OH ion in a water sample is found to be 2.5 x 10 -6 M pOH = – log₁₀ [OH – ] pOH = – 1og₁₀ [2.5 x 10 -6 ] = – log₁₀ [2.5] – log₁₀ [10 -6 ] = – 0.3979 – ( – 6) = – 0.3979 + 6 pOH = 5.6 Since pOH is less than 7, the solution is basic

Question 7. A lab assistant prepared a solution by adding a calculated quantity of HCl gas 25°C to get a solution with [H₃O + ] = 4 x 10 5 M. Is the solution neutral (or) acidic (or) basic. Answer: [H₃O + ] = 4 x M pH = – log₁₀ [H₃O + ] pH = – 1og₁₀[4 x 10 5 ] pH = – log₁₀ – log₁₀ [10 -5 ] pH = – 0.6020 – ( – 5) = – 0.6020 + 5 pH = 4.398 Therefore, the solution is acidic.

Question 8. Calculate the pH of 0.04 M HNO₃ solution. Answer: Concentration of HNO₃ = 0.04M [H₃O + ] = 0.04 mol dm -3 pH = – 1og[H₃O + ] = – log (0.04) = – log(4 x 10 -2 ) = 2 – log4 = 2 – 0.6021 = 1.3979 = 1.40

Concern 9. Explain solubility equipment. Answer: Solubility product: It’s defined as the product of molar concentration of the fresh constituent ions, for every single raised for the energy of the stoichiometric coefficient during the good well-balanced harmony formula.