Acids eg formic acid and you will acetic acidic try partly ionised inside solution and just have lower K

Obtain the worth of solubility product away from molar solubility

2. Acids such as HCI, HNO₃ 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: 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 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) i.age., in the event the dilution grows because of the 100 moments (amount minimizes in one x ten -dos Meters to 1 x 10 -cuatro Yards), the brand new dissociation develops by ten times.

1. Buffer is a simple solution using its a mix of weak acid as well as conjugate base (or) a deep failing ft and its conjugate acid.
2. That it boundary services resists radical changes in the pH on inclusion regarding a small levels of acids (or) bases and this feature is named barrier action.
3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH₄O and NH₄Cl.

1. The new buffering element of a simple solution are going to be counted in terms from shield capabilities.
2. Barrier index ?, as a decimal way of measuring new boundary potential.
3. It is recognized as the number of gram alternatives out of acid or foot put into 1 litre of shield option to changes its pH of the 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 10. Exactly how was solubility product is familiar with choose this new precipitation out-of ions? When the product out of molar intensity of the brand new constituent ions we.elizabeth., ionic device exceeds the fresh solubility tool then your substance becomes 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. By this means, the fresh solubility device finds good for choose if or not an ionic material will get precipitated whenever solution containing the latest component ions is actually mixed.

Question eleven. Solubility can be computed off molar solubility.i.e., the utmost number of moles of the solute which might be dissolved in one single litre of services.

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- https://datingranking.net/escort-directory/milwaukee/ ] = ns K_sp = [X n+ ] m [Y m- ] n K_sp = (ms) m (ns) n K_sp = (m) m (n) n (s) m+n