Derive the value of solubility unit away from molar solubility

2. Acids such as HCI, HNO_{step step 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: 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 Montgomery AL eros escort 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.elizabeth., if dilution grows by the one hundred times (focus reduces from just one x 10 -2 Meters to at least one x 10 -4 Yards), the newest dissociation increases by ten moments.

1. Barrier try a remedy using its a mix of poor acidic and its conjugate foot (or) a deep failing foot as well as conjugate acid.
2. It buffer provider resists radical alterations in its pH up on inclusion regarding a small levels of acids (or) bases hence feature is known as barrier step.
3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH₄O and NH₄Cl.

1. The buffering function of an answer are going to be counted with regards to regarding shield capabilities.
2. Shield list ?, as the a decimal measure of the shield capacity.
3. It’s recognized as what number of gram equivalents away from acidic otherwise base set in step one litre of one’s buffer substitute for changes their 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. Exactly how try solubility product is accustomed choose the newest rain off ions? In the event the unit off molar concentration of the new constituent ions i.age., ionic equipment exceeds the solubility device then material 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.

step 3. By this way, the fresh new solubility tool discovers advantageous to choose whether a keen ionic compound gets precipitated when provider containing the fresh constituent ions was combined.

Matter eleven. Solubility is calculated of molar solubility.i.elizabeth., the utmost number of moles of the solute and this can be demolished in one litre of the 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