Acids such a greats formic acid and you will acetic acid was partly ionised in the solution and also have lower K
2. Acids such as HCI, HNO3 are almost completely? onised and hence they have high Ka value i.e., Ka for HCI at 25°C is 2 x ten 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 escort services Huntsville 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) i.elizabeth., if the dilution grows because of the one hundred moments (amount minimizes from a single x ten -2 Meters to at least one x 10 -cuatro Meters), the fresh new dissociation grows by the 10 times. Question 10. Just how is actually solubility product is used to determine the fresh precipitation off ions? If unit away from molar intensity of brand new component ions i.elizabeth., ionic tool is higher than the fresh solubility equipment then your substance will get 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. step 3. By this means, the latest solubility device discovers advantageous to determine whether an enthusiastic ionic compound becomes precipitated when service containing brand new component ions is combined. Matter eleven. Solubility are going to be determined out of molar solubility.we.e., the most level of moles of your own solute that may be demolished in one litre of one’s 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 1