Acids such as for instance formic acid and you will acetic acidic was partially ionised for the services as well as have lowest K
Derive the value of solubility product off molar solubility

2. Acids such as HCI, HNOstep step 3 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.

  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 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

(iv) we.elizabeth., in the event the dilution increases of the one hundred times (focus decreases from a single x 10 -2 Yards to a single x ten -cuatro Yards), the latest dissociation develops by 10 minutes.

  1. Boundary is a remedy having its a combination of weak acidic and its particular conjugate base (or) a failure legs and its conjugate acid.
  2. That it buffer provider resists drastic alterations in their pH abreast of introduction out-of a small quantities of acids (or) bases and this element is called shield step.
  3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing NH4O and NH4Cl.
  1. Brand new buffering feature of a solution should be counted in terms out-of barrier skill.
  2. Barrier index ?, because the a decimal way of measuring the brand new barrier capability.
  3. It is identified as the number of gram competitors out of acidic otherwise foot added to 1 litre of your own boundary substitute for changes their pH by the unity.
  4. ? = \(\frac escort services in Irvine < dB>< d(pH)>\). 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 10. How are solubility device is accustomed pick the fresh new rain out of ions? If the unit regarding molar intensity of the brand new constituent ions i.e., ionic tool is higher than the brand new solubility equipment then substance gets 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 3. Through this means, the brand new solubility tool finds out beneficial to determine whether or not an ionic compound will get precipitated when solution containing new component ions was combined.

Question eleven. Solubility would be determined off molar solubility.we.e., the most level of moles of solute which are often mixed in one single litre of your provider.

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