All Questions
Tagged with acid-base stoichiometry
47
questions
-1
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Why does vinegar and baking soda yield a far smaller volume of carbon dioxide gas than expected from calculation?
I experimented $\ce{CO2}$ generation by adding $\pu{4 g}$ of $25\%$ acetic acid water solution ($= \pu{1 g}$ of acetic acid) to $\pu{1.4 g}$ of pure baking soda. But this mixture yielded only $50\%$ ...
- 11
1
vote
2
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212
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Weight of sodium acetate required to get to certain pH
How many grams of $\ce{CH3COONa}$ we need to dissolve in $\pu{2.50 L}$ of water to get a solution with $\mathrm{pH} = 9.00$? Given $K_\mathrm{a}(\ce{CH3COOH}) = 1.75\cdot 10^{-5}$
Like solving a ...
- 305
2
votes
1
answer
2k
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Mixing Ca(OH)2 with Ca(NO3)2
Mixing $\ce{Ca(OH)_2}$ with $\pu{0.1 M}$ concentration of $\ce{Ca(NO3)2}$.
What will actually happen in the process? What I will see with my eyes? I know that $\ce{Ca^2+}$ cation with concentration ...
- 305
0
votes
1
answer
513
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Equivalent weight of the anhydrous acid [closed]
The cupric salt of a monobasic acid contains $3$ molecules of water of hydration per atom of $\ce{Cu}$. $\pu{1 g}$ of hydrated salt yielding on strong heating $\pu{0.3306 g}$ of $\ce{CuO}$. What is ...
- 167
1
vote
2
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390
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Neutralization of alkali solution by a mix of strong acids
$\pu{0.1 M}$ $\ce{HCl}$ and $\pu{0.2 M}$ $\ce{H2SO4}$ solutions are mixed in equal volume. This solution is diluted to double the volume. Find the volume of $\pu{0.1 M}$ $\ce{NaOH}$ which can be ...
- 141
0
votes
2
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452
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Calculation for buffer of pH 5 in sodium fluoride solution
From a journal:
$\pu{500 mL}$ of $\pu{50 ppm}$ fluoride at $\mathrm{pH}~5$ was prepared by mixing $\pu{0.0552 g}$ of sodium fluoride with $\pu{4.1057 g}$ of sodium acetate and $\pu{1.65 mL}$ of ...
- 23
1
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2
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What's the best way to interconvert between molarity and percentage?
Let's say that I want to convert concentration of an acid from % to molarity.
One way I found to do this is by using the following formula:
percentage = (molarity * molar mass) / 10.
Another way is ...
- 202
0
votes
1
answer
273
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Why doesn't hydroxide concentration equal concentration of carbonic acid and bicarbonate in a sodium bicarbonate solution?
To find the pOH of solution of $\pu{0.42 M}$ of $\ce{NaHCO3}$, given a $\ce{Kb}$ of $\pu{2.4E-8}$, my solution manual does the following steps:
Reaction is similar to : $\ce{HCO3- + H2O <=> ...
- 2,853
0
votes
0
answers
63
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Theoritical amount of Sodium Bicarbonate
I have to find the theoretical amount of sodium bicarbonate. The steps I followed in order to produce it are:
15g $\ce{NaCl}$ are added to a beaker. Then we add 87.5 mL of $\ce{NH3}(12M)$
We add 115g ...
user40808
3
votes
2
answers
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If there are multiple bases in solution and a strong acid is added, will both bases react?
For example, if there is one mole of carbonate ions ($\ce{CO3 ^2-}$) and sulfate ions ($\ce{SO4 ^2-}$) in a solution, and one mole of hydrochloric acid ($\ce{HCl}$) is added, will the one mole of ...
- 309
0
votes
1
answer
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Why is finding [H+] when given [OH-]= 3.57e-5 M wrong when I use -log[OH-]?
Question/Problem
I know that [H+][OH-]=$10^{-14}$
And the problem asks to find out the concentration of H+ when the concentration of OH- is $3.57*10^{-5}$M
By using the equation above i find that ...
- 300
2
votes
2
answers
10k
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Why is potassium phosphate KH2PO4 in this reaction?
This is a continuation of this question because the first thing that came in my mind is that why potassium phosphate in this reaction is $\ce{KH2PO4}$ and not $\ce{K3PO4}$?
In the wikipedia article ...
- 26.3k
1
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2
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How to balance the equation for the reaction of aluminum bicarbonate and hydrochloric acid?
Here's what I have to balance:
Solid aluminum bicarbonate and hydrochloric acid combine to form an aluminum chloride solution and carbonic acid.
Or in an equation form:
$$\ce{\text{__ } Al(...
- 614
2
votes
2
answers
2k
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Balanced equation for neutralisation reaction betweeen potassium carbonate and hydrochloric acid
They seem all correct to me, is there any reason why one reaction would happen over the other(s)?
$$\begin{align}
\ce{K2CO3 + 2 HCl &-> 2 KCl + H2CO3}\\[6pt]
\ce{K2CO3 + 2 HCl &-> 2 KCl ...
- 21
5
votes
1
answer
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Back titration of excess HCl used for dissolving limestone
$15.0~\mathrm{mL}$ of $1.4~\mathrm{M}\ \ce{HCl}$ was mixed with $1.00~\mathrm{g}$ of limestone (impure $\ce{CaCO3}$) until all the solid had dissolved. The solution was then transferred to a conical ...
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