`20 Ml` of `KOH` solution was titrated with `0.20 M H_(2)SO_(4)` solution in a conductivity cell. The data obtained were plotted to give the graphe shown below.
The concentration of the `KOH` solution was

100 ml of 0.3 M HCl solution is mixed with 200 ml of 0.3 M H_(2)SO_(4) solution. What is the molariyt of H^(+) in resultant solution ?

20 ml of 20 vol H_(2)O_(2) solution is diluted to 80 ml. The final volume strength of solution is

A 100 ml solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued after adding 30 ml NaOH solution. The remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is

"40 ml "(N)/(10)HCl solution is mixed with 60 ml of (N)/(20)KOH solution. The resulting mixture will be

100mL of " 0.1 M NaOH" solution is titrated with 100mL of "0.5 M "H_(2)SO_(4) solution. The pH of the resulting solution is : ( For H_(2)SO_(4), K_(a1)=10^(-2))

250 ml of 0.10 M K_2 SO_4 solution is mixed with 250 ml of 0.20 M KCI solution. The concentration of K^(+) ions in the resulting solution will be:

Volume of 0.1 M H_2SO_4 solution required to neutralize 20 ml of 0.2 M NaOH solution is :

Volume of 0.1 M H_2SO_4 solution required to neutralize 20 ml of 0.1 M NaOH solution is :

Find % (w/v) of 20% (w/w) H_(2)SO_(4) solution, if density of solution is 1.2 gram/ml.

20 mol of M//10 CH_(3)COOH solution is titrated with M//10 NaOH solution. After addition of 16 mL solution of NaOH . What is the pH of the solution (pK_(a) = 4.74)