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Why does the half-equivalence point of HCl in the case of titrating strong base into HCl not yield the pKa?
If I have 20 mL of 0.1 M HCl and I titrate it with 0.1 M NaOH, the equivalence point should be about at pH = 7, and 20 mL of titrant.
The pH of 0.1 M HCl is about -log(0.1) = 1, so the pH at the half-equivalence point has to be between 1 and 7. If this was a weak acid being titrated with a strong base then this point would yield the pKa.
However, for HCl the half-equivalence point doesn't yield the pKa. Lots of literature suggests that the pKa of HCl is less than 0.
Is this determined by titration somehow?
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4 AnswersChemistry2 years agoWhat is the limiting reactant, and why I am I getting a contradiction?
Consider the burning of hydrogen gas using oxygen to produce water:
Balanced: 2H2 + O2 ----> 2H2O
If it was somehow possible to put an equal number of H2 and O2 molecules in the system. So 2 of each,
I would get 2 molecules of H2O produced with 1 molecule of unreacted oxygen. Since oxygen is in excess then H2 must be the limiting reactant.
But if I start with 1 gram of each, from 1 gram H2 I get 9 grams H2O; whereas from 1 gram O2 I get 1.125 g H2O. Therefore, I get less product from O2 so oxygen should be the limiting reactant.
Calculations:
1 g H2 * (1 mole H2 / 2 g H2) * (2 moles H2O / 2 moles H2) * (18 grams H2O / 1 mole H2O) = 9 grams H2O
1 g O2 * (1 mole O2 / 32 g O2) * (2 moles H2O / 1 mole O2) * (18 g H2O / 1 mole H2O ) = 1.125 g H2O
2 molecules H2 * (2 molecules H2O / 2 molecule H2) = 2 molecules of H2O
2 molecule O2 * (2 molecules H2O / 1 molecule O2) = 4 molecules H2O.
Summary:
When I use an equal number of molecules of each reactant,
I get less molecules of H2O from H2 so I conclude H2 is the limiting reactant.
When I use an equal number of grams of each reactant, I get less mass (g) of H2O from O2 so I conclude O2 is the limiting reactant.
Did I somehow choose a mass in grams of O2 which is an exception to the rule?
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2 AnswersRock and Pop2 years agoHow do you integrate this Michaelis Menten equation: -d[S] / dt = {V* [S]} / {K + S}?
This is the Michealis-Menten Equation. Integration of it is useful for when the substrate concentration [S] is directly measurable like by Nuclear Magnetic Resonance, but estimating the rate of substrate consumption introduces some error.
-d[S] / dt = {V* [S]} / {K + [S]}
[S] = substrate concentration
V = maximum velocity
K = Michaelis constant
dt respects time.
Follows from: E + S ⇌ ES ---> E + P
1 AnswerMathematics2 years agoIs my understanding of volume:volume mixtures correct?
If someone asks me to produce 1000 mL of a 7:9 volume to volume dilution of limonene and water,
Is that 7 parts of limonene + 9 parts of water in my volumetric flask = 16 parts in my flask?
Therefore the volume of limonene (L) that I should add is (7 / 16) * 1000 mL = 437.5 mL of Limonene
The volume of water to add is (9 / 16) * 1000 mL = 562.5 mL
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So 562.5 mL of water added to 437.5 mL of limonene makes a 7:9 v/v mixture of limonene and water?
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