Acids eg formic acidic and you may acetic acidic was partly ionised for the provider and also have lower K

Get the value of solubility product of molar solubility

2. Acids such as HCI, HNO_{step three} are almost completely? onised and hence they have high K_a value i.e., K_a for HCI at 25°C is 2 x ten 6 .

cuatro. Acids with K_a value greater than ten are considered as strong acids and less than one considered as weak acids.

1. HClO₄, HCI, H₂SO₄ – are strong acids
2. NH₂ – , O 2- , H – – are strong bases
3. 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 K_a 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.age., in the event the dilution develops from the a hundred moments (quantity decreases from 1 x 10 -dos Meters to just one x ten -4 Yards), the brand new dissociation increases from the ten moments.

1. Boundary try a remedy having its a combination of weakened acidic and its conjugate legs (or) a failure base and its own conjugate acid.
2. It barrier services resists extreme changes in their pH through to inclusion from a little levels of acids (or) basics and escort services in Pembroke Pines this element is named buffer step.
3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH₄O and NH₄Cl.

1. The new buffering feature out-of a remedy is going to be mentioned when it comes off boundary capacity.
2. Shield directory ?, just like the a decimal way of measuring the brand new barrier strength.
3. It is recognized as just how many gram competitors out-of acidic otherwise ft put into step one litre of one’s boundary option to alter its pH by unity.
4. ? = \(\frac < dB>< d(pH)>\). 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. How was solubility device is accustomed decide brand new precipitation from ions? If equipment away from molar intensity of the brand new component ions i.age., ionic product exceeds new solubility tool then the material will get precipitated.

2. When the ionic Product > K_sp precipitation will occur and the solution is super saturated. ionic Product < K_sp no precipitation and the solution is unsaturated. ionic Product = K_sp equilibrium exist and the solution ?s saturated.

3. From this way, the fresh solubility product discovers useful to pick whether a keen ionic material gets precipitated when solution that contain the component ions is combined.

Matter 11. Solubility should be determined away from molar solubility.we.elizabeth., the most level of moles of the solute that can be mixed in a single litre of your own service.

3. From the above stoichiometrically balanced equation, it is clear that I mole of X_m Y_n(s) dissociated to furnish ‘m’ moles of x and ‘n’ moles of Y. If’s’ is the molar solubility of X_m Y_nthen Answer: [X n+ ] = ms and [Y m- ] = ns K_sp = [X n+ ] m [Y m- ] n K_sp = (ms) m (ns) n K_sp = (m) m (n) n (s) m+n