Acids instance formic acid and acetic acid was partially ionised when you look at the solution as well as have lowest K

Obtain the worth of solubility device out of molar solubility

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

4. 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 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

(iv) i.age., if the dilution develops by the a hundred moments (attention minimizes from just one x 10 -dos Meters to at least one x ten -cuatro Meters), the brand new dissociation expands from the 10 times.

1. Shield is a simple solution using its a combination of poor acid and its particular conjugate foot (or) a failure legs and its own conjugate acid.
2. That it boundary services resists extreme changes in its pH on inclusion regarding a little levels of acids (or) angles and that feature is known as buffer action.
3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH₄O and NH₄Cl.

1. New buffering element off a solution is mentioned with regards to out-of boundary strength.
2. Shield list ?, as the a decimal measure of the brand new shield ability.
3. It is defined as just how many gram equivalents away from acidic or feet put in 1 litre of shield choice to changes their 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.

Question ten. How try solubility device is familiar with determine this new precipitation from ions? When the tool regarding molar concentration of the latest constituent ions i.age., ionic tool exceeds the fresh solubility product 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.

step three. From this method, the new solubility unit finds advantageous to determine whether an ionic substance will get precipitated when solution containing brand new component ions try mixed.

Concern 11. Solubility are going to be computed out of molar solubility.we.age., the most level of moles of solute which might be dissolved in one single litre of 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