The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is
(1)9.30
(2)7.30
(3)10.30
(4)8.30{{plese explen}}

Question

The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is
(1)9.30
(2)7.30
(3)10.30
(4)8.30{{*plese explen*}}

Pankaj · Answer

Hello Studnt,
This problem can be solved usingHenderson's Equation as follows
pOH = pKb+log[KCN]/[HCN]
Here we know that the number of moles of KCN and HCN are equal in the solution so,[KCN] =[HCN] and
We get,
pOH = pKb+log 1
or pOH =pKb= 4.7
Now,
pH =14-pOH = 14-4.7 = 9.30

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The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is (1)9.30 (2)7.30 (3)10.30 (4)8.30{{plese explen}}

The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is
(1)9.30
(2)7.30
(3)10.30
(4)8.30{{plese explen}}

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The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is (1)9.30 (2)7.30 (3)10.30 (4)8.30{{*plese explen*}}

The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is (1)9.30 (2)7.30 (3)10.30 (4)8.30{{*plese explen*}}

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The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is (1)9.30 (2)7.30 (3)10.30 (4)8.30{{plese explen}}

The PKb of cycanide ion is 4.7.The pH of solution prepared by mixing 2.5moles of KCN and 2.5moles of HCN in water and making total volume upto 500ml is
(1)9.30
(2)7.30
(3)10.30
(4)8.30{{plese explen}}