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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:
Explanation:
From the given information:
mass of silver chloride AgCl = 11.89 g
molar mass of AgCl = 143.37 g/mol
We know that:
number of moles = mass/molar mass
∴
number of moles of AgCl = 11.89 g/ 143.37 g/mol
number of moles of AgCl = 0.0829 mol
The chemical equation for the mineral called trona is:
when being reacted with hydrochloric acid, we have:
One mole of NaCl formed from one mole of trona sample = 0.0829 moles of AgCl
i.e. 0.0829 moles of NaCl can be formed from AgCl
mass of trona sample = number of moles × molar mass
mass of trona sample = 0.0829 × 226
mass of trona sample = 18.735 g
The mass in the percentage of NaHCO₃ = mass of NaHCO₃/ mass of trona
The mass in the percentage of NaHCO₃ = 6.93/18.735
The mass in the percentage of NaHCO₃ = 0.36989
The mass in the percentage of NaHCO₃ = 36.99%
Nonetheless, a 6.78 g samples manufactured from sodium carbonate in pure 100%
∴
6.78 g sample manufactured from Na₂CO₃ is purer.