Home
Class 12
CHEMISTRY
50 mL of 0.1 M NaOH is added to 60 mL of...

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

`3.1`

B

`5.5`

C

`4.1`

D

`6.5`

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • IONIC EQUILIBRIUM

    RESONANCE ENGLISH|Exercise PART-2 Single and double value integer type|9 Videos

  • IONIC EQUILIBRIUM

    RESONANCE ENGLISH|Exercise partIII one or more than one options correct type|10 Videos

  • IONIC EQUILIBRIUM

    RESONANCE ENGLISH|Exercise Exercise 1 part 2 only one option correct type|25 Videos

  • HYDROCARBON

    RESONANCE ENGLISH|Exercise ORGANIC CHEMISTRY(Hydrocarbon)|50 Videos

  • IUPAC NOMENCLATURE & STRUCTURAL ISOMERISM

    RESONANCE ENGLISH|Exercise Advanced Level Problems Part-5|2 Videos

Similar Questions

Explore conceptually related problems

Approximate p.H,0.1M aqueous H_(2)S solution when K_(1) and K_(2) for H_(2)S at 25^(@)C are 1xx10^(-7) and 1.3xx10^(-13) respectively:-

K_(1) and K_(2)\ of H_(2) S are 10^(-3) and 10^(-12) respectively what is [S^(-2)] in 0.1 M H_(2) S buffered at p(H) =2

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

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 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 ?

If 50ml of 0.2 M KOH is added to 40 ml of 0.05 M HCOOH, the pH of the resulting solution is ( K_(a)=1.8xx10^(-4) )

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 :

Calculate the equilibrium constants for the reactions with water of H_(2)PO_(4)^(Theta), HPO_(4)^(2-) , and PO_(4)^(3-) as ase. Comparing the relative values of two equilibrium constants of H_(2)PO_(4)^(Theta) with water, deduce whether solutions of this ion in water are acidic or base, Deduce whether solutions of HPO_(4)^(2-) are acidic or bases. Given K_(1),K_(2) ,and K_(3) for H_(3)PO_(4) are 7.1 xx 10^(-3), 6.3 xx 10^(-8) , and 4.5 xx 10^(-13) respectively.

The addition of NaH_(2)PO_(4) to 0.1M H_(3)PO_(4) will cuase

For a polyprotic acid, H_(3)PO_(4) its three dissociation constanst K_(1),K_(2) and K_(3) are in the order