An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings,
1 answer:
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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
The equation for dissociation of lactic acid is:
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]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bconjugated%20base%5D%7D%7B%5BAcid%5D%7D%20)
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86
The number of moles of lactic acid and lactate are as follows:
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
[CH₃CH(OH)COO⁻] = n (lactate) / 660 ml = 0.45 M
pH = 3.86 + log
= 3.86 + 0.191 = 4.05
So the pH of given solution is 4.05
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
The equation for dissociation of lactic acid is:
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
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86
The number of moles of lactic acid and lactate are as follows:
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
[CH₃CH(OH)COO⁻] = n (lactate) / 660 ml = 0.45 M
pH = 3.86 + log
So the pH of given solution is 4.05
Answer:
The volume of NaOH required is - 0.01 L
Explanation:
At equivalence point ,
Moles of = Moles of NaOH
Considering :-
Given that:
So,
<u>The volume of NaOH required is - 0.01 L</u>