2. Acids such as HCI, HNO3 are almost completely? onised and hence they have high Ka value i.e., Ka for HCI at 25°C is 2 x ten 6 .
cuatro. Acids with Ka value greater than ten are considered as strong acids and less than one considered as weak acids.
- HClO4, HCI, H2SO4 – are strong acids
- NH2 – , O 2- , H – – are strong bases
- 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 Ka 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., if the dilution expands by 100 minutes (quantity decreases from just one x ten -2 M to at least one x 10 -4 M), Salem escort reviews the dissociation develops because of the 10 times.
- Shield is actually a remedy using its a mix of weakened acidic and its particular conjugate feet (or) a weak foot and its own conjugate acidic.
- Which shield service resists drastic alterations in its pH upon addition from a little quantities of acids (or) basics and that function is known as shield step.
- Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH4O and NH4Cl.
- The new buffering function out-of a remedy will be mentioned in terms off buffer strength.
- Shield list ?, given that a decimal way of measuring the buffer capabilities.
- It is recognized as just how many gram equivalents from acidic or foot set in 1 litre of your buffer choice to transform its pH from the unity.
- ? = \(\frac < dB>< d(pH)>\). 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 are solubility product is used to determine the latest precipitation out of ions? In the event that device away from molar concentration of the constituent ions we.age., ionic unit exceeds new solubility device then the compound becomes precipitated.
2. When the ionic Product > Ksp precipitation will occur and the solution is super saturated. ionic Product < Ksp no precipitation and the solution is unsaturated. ionic Product = Ksp equilibrium exist and the solution ?s saturated.
step three. Through this ways, the latest solubility product finds good for pick whether or not a keen ionic compound gets precipitated whenever solution that contain the fresh new constituent ions try mixed.
Question eleven. Solubility is going to be calculated out-of molar solubility.we.elizabeth., the utmost quantity of moles of the solute that can be dissolved in a single litre of services.
3. From the above stoichiometrically balanced equation, it is clear that I mole of Xm Yn(s) dissociated to furnish ‘m’ moles of x and ‘n’ moles of Y. If’s’ is the molar solubility of Xm Ynthen Answer: [X n+ ] = ms and [Y m- ] = ns Ksp = [X n+ ] m [Y m- ] n Ksp = (ms) m (ns) n Ksp = (m) m (n) n (s) m+n