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3 years ago

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Consider a buffer solution containing CH3COOH and CH3COO-, with an equilibrium represented by: CH3COOH(aq) + H2O (l) ←----→ H3O+

(aq) + CH3COO- (aq) Describe what occurs if a strong acid such as HNO3 is added to the system, including an explanation of the direction of the equilibrium shift. Describe what occurs if a strong base such as KOH is added, including an explanation of the direction of the equilibrium shift.

1 answer:

Irina-Kira [14]3 years ago

5 0

Answer:

Here's what I get.

Explanation:

(a) The buffer equilibrium

The equation for the buffer equilibrium is

$\rm CH_{3}COOH(aq) + H$_{2}$O(l) $\, \rightleftharpoons \,$ CH$_{3}$COO$^{-}$(aq) + H$_{3}$O$^{+}$(aq)$

(b) Addition of acid

If you add a strong acid like HNO₃, you are increasing the concentration of hydronium ion.

Per Le Châtelier's Principle, the system will respond in such a way as to decrease the concentration of hydronium ion.

The position of equilibrium will shift to the left.

(c) Addition of base.

If you add a strong base like KOH, The hydroxide ions will react with the hydronium ions to form water.

The concentration of hydronium ions will decrease.

Per Le Châtelier's Principle, the system will respond in such a way as to increase the concentration of hydronium ions.

The position of equilibrium will shift to the right.

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The activation energy is $=8.1\,kcal\,mol^{-1}$

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The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

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Volume "V" = $10dm^{3}$

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Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

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$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

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