The `pK_(b)` of `CN^(Theta)` is `4.7`. The `pH` is solution prepared by mixing `2.5 mol` of `2.5 mol` of `KCN` of `2.5 mol` of `HCN` in water and making the total volume upto `500mL` is

The pK_(a) of HCN is 9.30. The pH of a solutin prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making up the total volume to 500 mL is

The pKa of HCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making up the total volume of 500 ml is :

The dissociation constant of HCN is 5 xx 10^(-10) . The pH of the solution prepared by mixing 1.5 mole of HCN and 0.15 moles of KCN in water and making up the total volume to 0.5 dm^(3) is

The dissociation constant of HCN is 5 xx 10^(-10) . The pH of the solution prepared by mixing 1.5 mole of HCN and 0.15 moles of KCN in water and making up the total volume to 0.5 dm^(3) is

Find the pH of a solution prepared by mixing 25ml of a 0.5M solution of HCI,10ml of a 0.5M solution of NaOH and 15 ml of water :-

Find the pH of a solution prepared by mixing 25ml of a 0.5M solution of HCI,10ml of a 0.5M solution of NaOH and 15 ml of water :-

Find the pH of solution prepared by mixing 25 ml of a 0.5 M solution of HCl, 10 ml of a 0.5 M solution of NaOH and 15 ml of water

A solution of NaOH is prepared by dissolving 1.5 g of the base in 500 mL of water. Calculate the pH of the solution.

If pK_(a) for CN^(-) at 25^(@)C is 4.7, the pH of 0.5 M aqueous NaCN solution is