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3 years ago

Determine the percent ionization of a 0.145 m hcn solution.

1 answer:

5 0

When Ka of HCN = 6.2x10^-10

HCN H3O+ CN-
initial 0.145 0 0
change -X +X +X
final (0.145-X) +X +X

Ka= [H3O+][CN]- / [HCN]
by substitution:
we can assume [HCN] = 0.145
6.2x10^-10 = X*X / (0.145) by solving this equation,
X = 9.5 x 10^-6
∴ [ H3O+] = 9.5 X 10^-6
∴percent ionization = [H3O+]/[HCN] *100
= 9.5X10^-6 / 0.145 *100
= 0.0066 %

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Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

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n = number of moles of gas

w = mass of ammonia gas = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = $27^oC=273+27=300K$

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R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

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$w=5.28g$

Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

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