The molecular mass of bromophenol blue is 692.0. Find the atomic mass for copper from a periodic table.
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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
This is an incomplete question, here is a complete question.
Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:
what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.
Answer : The volume of hydrogen gas that will be collected is 1.85 L
Explanation :
First we have to calculate the number of moles of aluminium.
Given mass of aluminium = 1.35 g
Molar mass of aluminium = 27 g/mol
The given chemical reaction is:
As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.
Thus, aluminium is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
2 moles of aluminium produces 3 moles of hydrogen gas
So, 0.005 moles of aluminium will produce = of hydrogen gas
Now we have to calculate the mass of helium gas by using ideal gas equation.
PV = nRT
where,
P = Pressure of hydrogen gas = 743 Torr
V = Volume of the helium gas = ?
n = number of moles of hydrogen gas = 0.075 mol
R = Gas constant =
T = Temperature of hydrogen gas =
Now put all the given values in above equation, we get:
Hence, the volume of hydrogen gas that will be collected is 1.85 L