Let’s think about the stabilities of one’s conjugate angles ClO

Let’s think about the stabilities of one’s conjugate angles ClO

Such anions are normalized so you’re able to higher the amount, it has got smaller appeal getting proton and therefore, usually become poor feet. Thus, the brand new relevant acidic will be strongest while the weak conjugate base has good acid and you can a strong conjugate legs enjoys poor acidic.

Question 5

This means ClOcuatro – 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, navigate to website [Cu(H2O)six] 2+ (aq) + 4NHstep 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. 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 new ammonia replaces five of your own liquids molecules within copper giving 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 nine. Establish solubility unit. Answer: Solubility tool: It is identified as the merchandise of molar concentration of this new component ions, each raised toward energy of their stoichiometric coefficient during the an excellent balanced balance equation.