Home
Class 11
CHEMISTRY
Calculate pH a) NaH(2)PO(4) b) Na(2)HPO(...

Calculate pH a) `NaH_(2)PO_(4)` b) `Na_(2)HPO_(4)` respectively, for `H_(3)PO_(4) pKa_(1) = 2.25, pKa_(2) = 7.20, pKa_(3) = 12.37`)

A

`9.78,4.68`

B

`4.68,4.68`

C

`9.78,9.78`

D

`4.68,9.78`

Text Solution

Verified by Experts

The correct Answer is:
D
`pH` of `H_(2)PO_(4)^(-) =- (pKa_(1) + pKa_(2))/(2)`
`pH` of `HPO_(4)^(-2) = (pKa_(2) + pKa_(3))/(2)`
Promotional Banner

Topper's Solved these Questions

  • IONIC EQUILIBRIUM

    NARAYNA|Exercise Level-V (H.W)|125 Videos

  • HYDROGEN & ITS COMPOUNDS

    NARAYNA|Exercise LEVEL-4|21 Videos

  • PERIODIC TABLE

    NARAYNA|Exercise All Questions|568 Videos

Similar Questions

Explore conceptually related problems

What will be the pH of a solution formed by mixing 10 ml 0.1 M NaH_(2)PO_(4) and 15 mL 0.1 M Na_(2)HPO_(4) ? ["Given: for "H_(3)PO_(4)Pk_(a_(1))=2.12, Pk_(a_(2))=7.2]

pH of 0.1 M Na_2HPO_4 and 0.2 M NaH_2PO_4 are respectively : ( pK_a for H_3PO_4 are 2.2,7.2,12.0)

pH of 0.1M Na_(2)HPO_(4) and 0.2M NaH_(2)PO_(4) are respectively: (pK_(a)"for" H_(3)PO_(4) are 2.12, 7.21 and 12.0 for respective dissociation to H_2PO_(4)^(-), HPO_(4)^(2-) and PO_(4)^(3-)) :

A buffer solution 0.04 M in Na_(2)HPO_(4) and 0.02 in Na_(3)PO_(4) is prepared. The electrolytic oxidation of 1.0 milli-mole of the organic compound RNHOH is carried out in 100 mL of the buffer. The reaction is RNHOH+H_(2)OrarrRNO_(2)+4H^(+)+4e^(-) The approximate pH of solution after the oxidation is complete is : [Given : for H_(3)PO_(4),pK_(a1)=2.2,pK_(a2)=7.20,pK_(a3)=12]

The anhydride of acid H_(3)PO_(4) and HPO_(3) are:

The chemical name of NaH_(2)PO_(4) is -

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

NARAYNA-IONIC EQUILIBRIUM-Level-VI (H.W)
  1. In the titration of a weak acid of known concentratin with a standard ...

    Text Solution

    |

  2. When phenolphthalein is used as the indicator in a titration of an HCl...

    Text Solution

    |

  3. During the titration of a weak base with a strong acid, one should use...

    Text Solution

    |

  4. What is the pH at the equivalence point in a titration of 0.2 M NH(3) ...

    Text Solution

    |

  5. A sample of 100 mL of 0.10 M weak acid, HA (K(a) = 1.0 xx 10^(-5)) is ...

    Text Solution

    |

  6. In the titration of 25.0 mL of 0.1 M aqueous acetic acid (K(a) = 1.8 x...

    Text Solution

    |

  7. The simulaneous solubility of AgCN (Ksp = 2.5 xx 10^(-16)) and AgCl (K...

    Text Solution

    |

  8. The salt Al(OH)(3) is involved in the following two equilibria, Al(O...

    Text Solution

    |

  9. An indicator is a weak acid and pH range of its colour is 3.1 to 4.5. ...

    Text Solution

    |

  10. AgBr((s)) + 2S(2)O(3(aq))^(2-) hArr [Ag(S(2)O(3))(2)]((aq))^(3-) + Br(...

    Text Solution

    |

  11. A 25.0 mL of 0.1 M weak acid HA is titrated with 0.1 M NaOH to the eq...

    Text Solution

    |

  12. The [H^(+)] in a solution containing 0.1 M HCOOH and 0.1 M HOCN [Ka fo...

    Text Solution

    |

  13. The P^(H) of pure water at 25^(0)C and 35^(0)C are 7 and 6 respectivel...

    Text Solution

    |

  14. Calculate the percentage hydrolysis in 0.003M aqueous solution of NaOC...

    Text Solution

    |

  15. The amount of (NH(4))(2)SO(4) in grams which must be added to 500 ml o...

    Text Solution

    |

  16. 100 ml of sample is removed from an aqueous solution saturated with Ca...

    Text Solution

    |

  17. An aqueous solution of metal chloride MCI(2)(0.05M) is saturated with ...

    Text Solution

    |

  18. The ratio of pH of solution (1) containing 1 mole of CH(3)COONa and 1 ...

    Text Solution

    |

  19. Calculate pH a) NaH(2)PO(4) b) Na(2)HPO(4) respectively, for H(3)PO(4)...

    Text Solution

    |

  20. At 25^(@)C, K(sp) for PbBr(2) is equal to 8xx10^(-5). If the salt is 8...

    Text Solution

    |