2. Acids such as HCI, HNOstep step three are almost completely? onised and hence they have high Ka value i.e., Ka for HCI at 25°C is 2 x 10 6 . cuatro. Acids with Ka value greater than ten are considered as strong acids and less than one considered as weak acids. Question 5. pH of a neutral solution is equal to 7. Prove it. in neutral solutions, the concentration of [H3O + ] as well as [OH – ] are equal to 1 x 10 -7 M at 25°C. 2. The pH of a neutral solution can be calculated by substituting this [H3O + ] concentration in the expression pH = – log10 [H3O + ] = – log10 [1 x 10 -7 ] = – ( – 7)log \(\frac < 1>< 2>\) = + 7 (l) = 7 Question 7. When the dilution increases by 100 times, the dissociation increases by 10 times. Justify this statement. Answer: (i). Let us consideran acid with Ka value 4 x 10 4 . We are calculating the degree of dissociation of that acid at two different concentration 1 x 10 -2 M and 1 x 10 -4 M using Ostwalds dilution law
(wev) we.elizabeth., in the event the dilution expands because of the a hundred moments (amount decrease from one x ten -dos Meters to at least one x 10 -4 M), the brand new dissociation develops of the 10 times. Matter 10. Just how is solubility device is always select the newest rain out of ions? If equipment away from molar intensity of the newest constituent ions i.age., ionic equipment exceeds new solubility unit then your compound becomes precipitated. 2. When the ionic Product > Ksp precipitation will occur and the solution is super saturated. ionic Product < Ksp no precipitation and the solution is unsaturated. ionic Product = Ksp equilibrium exist and the solution ?s saturated. 3. From this way, the solubility device finds out advantageous to select if or not an ionic compound will get precipitated when services containing the new component ions try combined. Concern eleven. Solubility is going to be calculated escort service in Omaha NE away from molar solubility.we.e., the maximum level of moles of one’s solute that may be dissolved in a single litre of your own provider. 3. From the above stoichiometrically balanced equation, it is clear that I mole of Xm Yn(s) dissociated to furnish ‘m’ moles of x and ‘n’ moles of Y. If’s’ is the molar solubility of Xm Ynthen Answer: [X n+ ] = ms and [Y m- ] = ns Ksp = [X n+ ] m [Y m- ] n Ksp = (ms) m (ns) n Ksp = (m) m (n) n (s) m+n
Answer: step one