`100 mL` of mixture of `NaOH` and `Na_(2)SO_(4)` is neutralised by `10 mL` of `0.5 M H_(2) SO_(4)`. Hence, in `100 mL` solution is

100 mL of mixture of NaOH and Na_(2)SO_(4) is neutralised by 10 mL of 0.5 M H_(2) SO_(4) . Hence,the amount of NaOH in 100 mL solution is

10ml of 1 M H_(2)SO_(4) will completely neutralise

250mL of 0.2M H_(2)SO_(4) is mixed with 100mL of 0.5M H_(2)SO_(4) . What is the molarity of the resulting solution?

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))

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))

How many ml of 1 (M) H_(2)SO_(4) is required neutralise 10 ml of 1 (M) NaOH solution?

100ml of 0.2 M H_(2)SO_(4) is added to 100 ml of 0.2 M NaOH . The resulting solution will be

100ml of 0.2 M H_(2)SO_(4) is added to 100 ml of 0.2 M NaOH . The resulting solution will be