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Acids such formic acidic and acetic acidic try partly ionised into the solution and have low K


Derive the value of solubility product out-of molar solubility

2. Acids such as HCI, HNOstep step three 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: 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 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.e., in the event that dilution expands of the 100 minutes (quantity decreases from 1 x ten -2 M to 1 x ten -cuatro Yards), brand new dissociation increases of the ten moments.

  1. Buffer is actually a remedy using its a mixture of weak acid and its own conjugate legs (or) a failure ft as well as conjugate acid.
  2. This barrier solution resists extreme alterations in the pH upon addition out of a tiny levels of acids (or) basics hence function is named barrier step. Loe edasi »

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