pH when 100 mL of 0.1 M `H_(3)PO_(4)` is titrated with 150 mL 0.1 m NaOH solution will be :

100ml of 0.1M H_(3)PO_(4) is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution K_(1)=10^(-3)M K_(2)=10^(-8)M K_(3)=10^(-13)M pH after HCI was added will be :

100ml of 0.1M H_(3)PO_(4) is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution K_(1)=10^(-3)M K_(2)=10^(-8)M K_(3)=10^(-13)M pH at 2nd equivalent point will be :

10mL of 0.1 M HCl solution reacts completely with 10 mL of 0.1 M NaOH solution. What will be the pH of the resulting solution?

25 mL of 0.107 M H_(3)PO_(4) was titrated with 0.115 M solution of NaOH to the end point identified by indicator bromocresol green.This required 23.1 mL . The titration was repeated using phenolphthalein as indicator. This time 25 mL of 0.107 M H_(3)PO_(4) reuired 46.2 mL of the 0.115 M NaOH . What is the coefficient of n in this equation for each reaction? H_(3)PO_(4)+n OH^(-)rarr nH_(2)O+[H_(3-n)PO_(4)]^(n-)

When 100 mL of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be :

When 100 mL of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be :

When 100 mL of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added 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))