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What is Kb for N2H4 if the pH of a 0.158M solution of N2H5Cl is 4.5?

Sagot :

Answer:

Kb[tex]= 1.58 * 10^{-6}[/tex]

Explanation:

The correct question is

What is Kb for N2H4 if the pH of a 0.158M solution of N2H5Cl is 4.5?

Solution-

N2H4Cl hydrolyses on addition of water

The reaction equation is as follows -

N2H4+ + H2O ----> N2H4 + H3O+

[tex]Ka = \frac{K_w}{K_b} = \frac{10^{-14}}{K_b}[/tex]

pH [tex]= 4.5[/tex]

[tex][H+] = 10^{-pH} = 10^{-4.5}[/tex]

[tex]K_b= \frac{10^{-14} }{K_b} = \frac{Y_2}{0.158} \\K_b= \frac{(10^{-4.5})^2}{0.158}\\=> Kb = 0.158 * 10-14 /(10-4.5)2[/tex]

 Kb[tex]= 1.58 * 10^{-6}[/tex]

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