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3 years ago

12

What is the formula for the conjugate base formed by the following acid HCN

1 answer:

meriva3 years ago

5 0

Answer:

HCN - Hydrogen cyanide The conjugate acid of CN- is HCN. HCN stands for hydrogen cyanide. The conjugate acids are a combination of a strong acid and a low base.

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How different isotopes of Helium are similar and different.

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In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated

lord [1]

Answer:

Explanation:

Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M

Then, the dissociation of 4-chlorobutanoic acid can be expressed as:

$\mathsf{C_3H_6ClCO_2H }$ ⇄ $\mathsf{C_3H_6ClCO_2^-}$ + $\mathsf{H^+}$

The ICE table can be computed as:

$\mathsf{C_3H_6ClCO_2H }$ ⇄ $\mathsf{C_3H_6ClCO_2^-}$ + $\mathsf{H^+}$

Initial 0.76 - -

Change -x +x +x

Equilibrium 0.76 - x x x

$K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}$

$K_a = \dfrac{[x] [x]}{ [0.76-x]}$

where:

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}$

however, the value of x is so negligible:

0.76 -x = 0.76

Then:

$3.02*10^{-5}*0.76 = x^2$

$x=\sqrt{3.02*10^{-5}*0.76 }$

x = 0.00479 M

∴

$x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=}$ 0.00479 M

$\mathsf{C_3H_6ClCO_2H }$ = (0.76 - 0.00479) M

= 0.75521 M

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= ( 0.00479 / 0.76) × 100

= 0.630 M

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Answer:

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