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Digiron [165]
3 years ago
13

Hno3, a strong acid, is added to shift the ag2co3 equilibrium (equation 7.6) to the right. explain why the shift occurs.

Chemistry
1 answer:
Deffense [45]3 years ago
7 0
Answer 1)  When a strong acid like HNO_{3} reacts with Ag_{2} CO_{3} usually the equilibrium shifts to the right because

As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change."  So, in the given reaction when HNO_{3} reacts with Ag_{2} CO_{3}  it generates carbon dioxide and water as a by product, if we are adding HNO_{3} it will remove some of the CO_{3} molecule from the reaction mixture, which then tends to shift the equilibrium towards right.

Answer 2) The same would be observed in this case, if we replace HNO_{3} with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.

As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing Ag^{+} and Cl^{-} ions from the solution.
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Muscle physiologists study the accumulation of lactic acid [ch3ch(oh)cooh] during exercise. food chemists study its occurrence i
Drupady [299]
The provided information are:
volume of 0.85 M lactic acid = 225 ml
volume of 0.68 M sodium lactate = 435 ml
Ka of the lactate buffer = 1.38 x 10⁻⁴
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CH₃CH(OH)COOH(aq) + H₂O ⇄ CH₃CH(OH)COO⁻(aq) + H₃O⁺(aq)
The pH of buffer is calculated from Henderson-Hasselbalch equation, which is:
pH = pKa + log \frac{[conjugated base]}{[Acid]}
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86 
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n (Lactic acid) = 225 ml x 0.85 mmol/ml = 191.25 mmol
n (Lactate) = 435 ml x 0.68 mmol/ml = 295.8 mmol
The number of moles of lactic acid and lactate in total volume of the solution:
[CH₃CH(OH)COOH] = n (lactic acid) / 660 ml = 191.25 mmol / 660 ml = 0.29 M
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pH = 3.86 + log \frac{0.45 M}{0.29 M} = 3.86 + 0.191 = 4.05
So the pH of given solution is 4.05
7 0
3 years ago
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