2. Acids such as HCI https://datingranking.net/escort-directory/los-angeles, HNOstep 3 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.
- HClO4, HCI, H2SO4 – are strong acids
- NH2 – , O 2- , H – – 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: 1
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 from the 100 minutes (concentration reduces from one x ten -dos M to just one x ten -cuatro Meters), the fresh new dissociation expands because of the ten times.
- Buffer are a simple solution having its a mixture of weak acidic and its own conjugate legs (or) a faltering legs and its own conjugate acid.
- Which buffer services resists radical alterations in the pH up on inclusion off a tiny levels of acids (or) angles and this element is named boundary step.
- Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH4O and NH4Cl.
- The fresh buffering element of an answer will likely be measured when it comes off boundary potential.
- Shield index ?, because the a quantitative measure of the latest buffer potential.
- It is identified as what amount of gram equivalents away from acidic or foot put in 1 litre of the barrier substitute for alter its pH by 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.
Matter ten. Just how is solubility product is regularly select the brand new precipitation from ions? If the product off molar intensity of brand new component ions i.e., ionic device exceeds the fresh new solubility equipment then the 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 way, the brand new solubility tool finds out useful to determine whether a keen ionic substance gets precipitated when service that contain this new component ions was mixed.
Question eleven. Solubility is going to be determined regarding molar solubility.we.e., the utmost amount of moles of the solute which may be mixed in one single litre of one’s service.
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