Acids particularly formic acid and you will acetic acidic is partially ionised inside service as well as have lower K
2. Acids such as HCI, HNOstep step 3 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.
- HClO4, HCI, H2SO4 – are strong acids
- NH2 – , O 2- , H escort in Gilbert – – are strong bases
- HNO2, HF, CH3COOH are 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
Answer: step one
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
(iv) we.e., when the dilution grows of the one hundred moments (amount reduces from 1 x 10 -dos Meters to at least one x ten -cuatro M), new dissociation grows because of the 10 moments.
- Boundary was an answer using its a mix of weakened acidic and its own conjugate legs (or) a faltering base and its conjugate acid.
- That it barrier services resists extreme changes in the pH through to inclusion off a small levels of acids (or) basics and this function is known as boundary action.
- Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH4O and NH4Cl.
- New buffering feature away from a remedy can be mentioned when it comes off shield skill.
- Barrier list ?, as the a quantitative way of measuring the fresh new boundary ability.
- It is identified as what number of gram equivalents away from acidic or legs put into step 1 litre of one’s shield choice to changes its pH because of the unity.
- ? = \(\frac < dB>< d(pH)>\). dB = number of gram equivalents of acid / base added to one litre of buffer solution. d(pH) = The change in the pH after the addition of acid / base.
Concern ten. How is actually solubility product is accustomed select new precipitation of ions? When the unit of molar concentration of this new constituent ions we.e., ionic unit is higher than the fresh solubility product then substance 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.
step three. Through this method, the fresh solubility product discovers good for select if or not an enthusiastic ionic substance gets precipitated when provider that contain brand new component ions are combined.
Concern eleven. Solubility are going to be determined of molar solubility.we.age., the maximum amount of moles of your own solute that can be demolished in one litre of solution.
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