Obtain the value of solubility tool off molar solubility

2. Acids such as HCI, HNO_{step 3} 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) we.age., in the event the dilution increases because of the a hundred minutes (amount minimizes from x 10 -dos M to just one x ten -cuatro Meters), the dissociation increases because of the 10 minutes.

1. Buffer is a simple solution which consists of a variety of weak acid and its particular conjugate feet (or) a failing base as well as conjugate acid.
2. This buffer solution resists extreme changes in the pH through to introduction regarding a tiny amounts of acids (or) bases and this function is known as shield action.
3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH₄O and NH₄Cl.

1. The brand new buffering feature out of a simple solution might be counted when it comes out of buffer potential.
2. Shield list ?, because a decimal way of measuring the newest barrier capability.
3. It is recognized as what number of gram counterparts of acid otherwise foot put into step one litre of one’s barrier solution to alter its pH from 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 ten. How are solubility product is accustomed choose the fresh new precipitation out of ions? In the event that tool away from molar intensity of the latest component ions i.e., ionic unit is higher than the brand new solubility device then substance 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. By this means, this new solubility device finds out useful to select if or not an enthusiastic ionic material will get precipitated when provider containing the latest component ions was blended.

Question eleven. Solubility is going to be computed regarding molar solubility.we.e., the utmost level of moles of your own solute that may be demolished in one litre of your solution.

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