60ml of 0.1M HCl is mixed with 40ml of 0.05M `H_(2)SO_(4)` .The "pH" value of the resulting solution is

10 ml of M/(200) H_(2)SO_(4) is mixed with 40 ml of M/200 H_(2)SO_(4) . The pH of the resulting solution is

100ml of 0.3 N HCl is mixed with 200ml of 0.6 N H_(2)SO_(4) . The final normality of the resulting solution will be -

100 ml of 0.15 M HCl is mixed with 100 ml of 0.05 M HCl, is the pH 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))

20 mL of 0.5 M HCl is mixed with 30 mL of 0.3 M HCl, the molarity of the resulting solution is :

100 ml of 0.3 M HCl is mixed with 200 ml of 0.3 M H_(2)SO_(4). Calculate the final molarity of the H^(+) ions in the resulting solution. Fill your answer by multiplying with 10.

5 mL of 0.4 N NaOH is mixed with 20 mL of 0.1 N HCl. The pH of the resulting solution will be

When 100 ml of 1 M NaOH solution is mixed with 10 ml of 10 M H_(2)SO_(4) , the resulting mixture will be