Acids like formic acid and you may acetic acid is partly ionised within the service and now have reasonable K

Get Rialto escort the value of solubility product 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: 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.elizabeth., if the dilution grows by 100 times (attention reduces from just one x 10 -dos Yards to at least one x 10 -cuatro Meters), the fresh dissociation increases of the ten moments.

1. Buffer is actually a solution using its a variety of weak acid and its particular conjugate foot (or) a faltering feet as well as conjugate acid.
2. So it boundary services resists drastic alterations in the pH on introduction of a tiny quantities of acids (or) bases and this ability is named barrier 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 simple solution will likely be mentioned when it comes of boundary skill.
2. Shield index ?, because the a decimal measure of the fresh boundary skill.
3. It is recognized as what number of gram equivalents out-of acid or legs added to step 1 litre of barrier option to transform the 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.

Concern ten. Exactly how was solubility device is always pick the latest precipitation away from ions? If product away from molar concentration of the latest component ions we.elizabeth., ionic device is higher than the fresh new solubility product 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.

step 3. From this method, this new solubility tool finds beneficial to choose whether or not a keen ionic substance will get precipitated whenever service containing the new component ions was combined.

Matter eleven. Solubility are going to be computed off molar solubility.i.elizabeth., the utmost number of moles of the solute that is certainly mixed in one single litre of your own 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- ] = ns K_sp = [X n+ ] m [Y m- ] n K_sp = (ms) m (ns) n K_sp = (m) m (n) n (s) m+n