19,000+ solved questions for JEE Advanced, JEE Mains, NEET & IChO — with answers and expert explanations.
The pH of 0.5 M aqueous solution of HF (Ka = 2 × 10–4) is
In decimolar solution, CH3COOH is ionised to the extent of 1.3%. If log 1.3 = 0.11, what is the pH of the solution ?
At a certain temperature the value of pKw is 13.4 and the measured pH of solution is 7. The solution is
What is the pH of a 0.015 M Ba(OH)2 solution ?
An acid solution of pH = 6 is diluted hundred times. The pH the solution becomes :
A is an aqueous acid; B is an aqueous base. They are diluted separately, then
The pH of a 0.1 M aqueous solution of a weak acid (HA) is 3. What is its degree of dissociation ?
Four grams of NaOH solid are dissolved in just enough water to make 1 litre of solution. What is the [H+] of the solution ?
The pOH value of a solution whose hydroxide ion concentration is 6.2 × 10–9mol/litre is
4g of NaOH are put into 10 litres of water. The pH of the resulting solution will be
The pH of solutions A, B, C, D are respectively 9.5, 2.5, 3.5, 5.5, The most acidic solution is
] ] In option B, it is a base salt. In option C, it is a weak acid salt. In option D, it is a strong acid
Which of the following solutions will have the highest pH value ?
A soda water bottle has pH
The term pH comes from
w 2H O H O OH ,K 1 10 x at 25ºC hence Ka is : (a) 1 10 x (b) 5.55 10 x (c) 18 10 x (d) 1.00 10 x
The concentration of [H+] and concentration of [OH–] of a 0.1 aqueous solution of 2% ionised weak monoprotic acid is [ionic product of water = 1×10–14]
At 90oC, pure water has H3O+ ion concentration of 10-6 mol/L. The Kw at 90oC is
The hydrogen ion concentration of a 10–8 M HCl aq. solution at 298K (Kw = 10–14) is (a) 1.0 10 M x (b) 1.10 10 M x (c) 9.525 10 M x (d) 1.0 10 M x
The concentration of water molecules in pure water at 298 K is
For a pure water,
A monoprotic acid in 1.00 M solution is 0.01% ionized. The dissociation constant of this acid is (a) 1 10 x (b) 1 10 x (c) 1 10 x (d) 1 10 x
At infinite dilution, the percentage ionisation for both strong and weak electrolytes is
Concentration CN – in 0.1 M HCN is [Ka = 4 × 10 –10]
For a weak acid HA, Ostwald’s dilution law is represented by the equation (a) a c K 2 (b) a c K (c) a K c c (d) a c K