All Questions
8
questions
1
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0
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Equivalence point pH derivation of a tribasic acid
Considering $\pu{100 mL} \ \pu{0.1 M} \ \ce{H3A}$ and $\pu{0.1 M} \ \ce{NaOH}$ titration curve:
I understood the half equivalence points, but couldn't understand the reason why $\mathrm{pH} = \frac{1}...
-1
votes
1
answer
161
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Finding a Mistake Calculating the pH of Ammonium Acetate
I'm trying to calculate the $\mathrm{pH}$ of a $\pu{0.10 M}$ solution of ammonium acetate, and I am struggling to find what mistake I am making within the following calculations:
We know that $\dfrac{[...
4
votes
3
answers
253
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Is pK_{In} = pH at equivalence point still true for weak basic organic indicators?
I am reading Ostwald Theory of Titration which says that indicators are organic weak acids or bases.
To prove the relation that $\mathrm{p}K_\mathrm{In} = \mathrm{pH}$, the textbook uses the following ...
3
votes
2
answers
821
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Calculating pH of rain water
The concentration of $\ce{SO2}$ in the atmosphere over a city on a certain day is $\pu{10 ppm}$. Given that solubility of $\ce{SO2}$ in water is $\pu{1.3653 mol L-1}$ and $\mathrm{p}K_\mathrm{a}$ of $\...
2
votes
1
answer
3k
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Calculating molecular weight of an acid based on given values of mass, volume, pH and pKa
This is what I thought was an easy problem but according to a given answer I am doing it wrong:
$\pu{11 mg}$ of a monoprotic acid was dissolved in $\pu{0.5 l}$ of water. The resulting $\mathrm{pH}$ ...
2
votes
5
answers
22k
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Why does a buffer work best at the pH closest to its pKa?
I think I'm having trouble understanding exactly how the $\mathrm{p}K_\mathrm{a}$ relates here on a conceptual level. For example I know that carbonic acid works somewhat well as a buffer at $\mathrm{...
2
votes
2
answers
37k
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Titration of H3PO4 with NaOH
I came across an ionic equilibrium problem stating:
Find the pH when 150 ml 1 M $\ce{NaOH}$ has been added to 100 ml 1 M $\ce{H3PO4}$.
I'm stuck with this question. What I know:
$$\ce{H3PO4 + NaOH ...
3
votes
2
answers
10k
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About pH of an aqueous solution of SO2
Probably we can have an aqueous solution of $\ce{SO2}$ by dissolving it in water, because we would have an equilibrium between $\ce{SO2(g)}$ and $\ce{SO2(aq)}$:
$$\ce{SO2(g) <=> SO2(aq)}$$
How ...