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2 years ago

The ph of 0.255 m hcn is 4.95. What is the value of ka for hydrocyanic acid?

Chemistry

1 answer:

Lera25 [3.4K]2 years ago

5 0

Answer:

Ka = 4.9 X 10^-9

Explanation:

when the pH is 4.95

the [H+] = 10^-4.95

[H+] = 1.122 e-5

since the equilibrium equation is:

HCN <=> 1 mole H+ & 1 mole CN-

then

[H+] = [CN-]

when

HCN <=> H+ & CN-

Ka = [ H+] [CN-] / [HCN]

Ka = [ 1.122 e-5] [ 1.122 e-5] / [0.255]

Ka = 1.256 e-10 / 0.255

Ka = 4.94 e-9

which rounds off to 2 sig figs into

Ka = 4.9 X 10^-9

======================

p.s. it was the pH pf 4.95 that limited the calculations to 2 sig figs

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Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.

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b) 213.8 °C= 416.84°F

c) 180 °C= 453.15 °K

d) 315 °K= 107.33 °F

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f) 0 °K= -459.67 °F

Explanation:

Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:

Fahrenheit a Celsius: $C=\frac{F-32}{1.8}$

Celsius a Fahrenheit: °F= °C*1.8 + 32
Celsius a Kelvin: °K= °C + 273.15
Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
Fahrenheit a Kelvin: $K=\frac{F-32}{1.8} + 273.15$

Entonces se obtiene:

a) 72 °F= $\frac{72-32}{1.8}$ =22.22 °C

b) 213.8 °C= 213.8*1.8 + 32= 416.84°F

c) 180 °C= 180°C + 273.15= 453.15 °K

d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F

e) 1750 °F= $\frac{1750-32}{1.8} + 273.15$ = 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

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