Such anions is normalized so you can higher the quantity, it offers minimal destination to have proton and that, tend to become poor foot. Consequently, the fresh new corresponding acid could well be most effective due to the fact weakened conjugate base enjoys strong acidic and a strong conjugate foot enjoys poor acid.

Question 5

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)₆] 2+ (aq) + 4NH_{step 3} (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 https://datingranking.net/escort-directory/orange/ 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 ammonia replaces four of your own drinking water particles around the copper supply 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

Matter nine. Describe solubility tool. Answer: Solubility product: It is identified as the item of the molar intensity of the component ions, per elevated toward electricity of its stoichiometric coefficient for the an effective balanced equilibrium formula.