50 mL of 0.1 M NaOH is added to 60 mL of 0.15 M `H_(3)PO_(4)` solution `(K_(1), K_(2) " and " K_(3) " for " H_(3)PO_(4) " are " 10^(-3), 10^(-8) " and " 10^(-13)` respectively). The pH of the mixture would be about (log 2=0.3) :

A solution containing a weak acid and its conjugate base acts as an acidic buffer. In a buffer the dissociation of the weak acid is suppressed by its conjugate base. (K_(1),K_(2), and K_(3) of H_(3)PO_(4) , are 10^(-4), 10^(-8), 10^(-13) respectively What is the pH of a solution obtained by mixing 100ml of 0.1 MH_(3) PO_(4) , and 150ml of 0.IM NaOH

A solution containing a weak acid and its conjugate acts as an acidic buffer. In a buffer the dissociation of the well acid is suppressed by its conjugate base. (K_(1),K_(2), and K_(3), of H_(3)PO_(4) , are 10_(-4), 10^(-4), 10^(-13) respectively Which of the following volume of 0.1 M NaOH added to 100 mlof 0.1M H_(3),PO_(4) , does not form a buffer

100ml 0.1M H_(3)PO_(4) is mixed with 50ml of 0.1M NaOH. What is the pH of the resultant solution? (Successive dissociation constant of H_(3) PO_(4) "are" 10^(-3) , 10^(-8) and 10^(-12) respectively ?

100ml 0.1M H_(3)PO_(4) is mixed with S0ml of 0.1M NaOH. What is the pH of the resultant solution? (Successive dissociation constant of H_(3) PO_(4) "are" 10^(-3) , 10^(-8) and 10^(-12) respectively ?

A solution containing a weak acid and its conjugate acts as an acidic buffer. In a buffer the dissociation of the well acid is suppressed by its conjugate base. (K_(1),K_(2), and K_(3), of H_(3)PO_(4) , are 10_(-4), 10^(-4), 10^(-13) respectively What is the P^(H) of a solution obtained by mixing 100ml of 0.1 MH(3) PO_(4) , and 150ml of 0.IM NaOH

100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOH solution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are " 10^(-3) , 10 ^(-6) and 10^(-14) respectively. What is the pH of the reaction mixture after adding 50 ml of NaOH ?

The pH of the resultant solution of 20 mL of 0.1 M H_(3)PO_(4) and 20 mL of 0.1 M Na_(3)PO_(4) is :

The pH of the resultant solution of 20 mL of 0.1 M H_(3)PO_(4) and 20 mL of 0.1 M Na_(3)PO_(4) is :

The volume of 0.025M Ca(OH)_(2) solution which can neutralise 100 ml of 10^(-4)M H_(3)PO_(4) is