All Questions
21
questions
-4
votes
1
answer
61
views
Is it possible to make acids and bases in amphiprotic solvents other than water [closed]
In such solvents, could there be chemicals which could donate/accept protons, and if so could their solutions display acidic/basic properties?
1
vote
0
answers
2k
views
Buffer solution of NaH2PO4 and Na2HPO4
We have this exercise without solutions
From a 0.2 M $\ce{NaH2PO4}$ solution and a 0.2 M $\ce{Na2HPO4}$ solution, a buffer
solution with pH = 6.8 is to be prepared. The total concentration of
$\ce{...
-1
votes
2
answers
1k
views
Why does a proton react with water to form hydronium ion and why does that ion has an overall positive charge? [closed]
Take this is an example:
$$\ce{HCl(aq) -> H^+(aq) + Cl^-(aq)}$$
and
$$\ce{HCl(aq) + H2O -> H3O+ + Cl^-(aq)}$$
Clearly the proton latches on to the water molecule but why? Also, why does the ...
0
votes
1
answer
213
views
Effect of mixing acids on solubility
Please help me clarify the title if you come up with a better way to put it (you certainly will).
My question is the following:
If you make a mix of equal parts of various acids (Bicarbonate, ...
-1
votes
1
answer
340
views
Hydrogen and Hydrogen ion
Is it correct that a hydrogen atom only has 1 electron and 1 proton since the atomic mass is 1?
Also, is the hydrogen ion composed of a single proton and no electrons?
In an ionization process we know ...
-1
votes
1
answer
98
views
Microscopic View of Adding Water to A Basic Solution
I confronted a problem, stating that we have a solution of potassium hydroxide. By adding extra water, when the volume of solution reaches 630 mL, the pH decreases by 0.5 units. How much extra water ...
1
vote
1
answer
87
views
Neutralisation Reaction Confusion
What will happen when $\pu{1 mol}$ of $\ce{H2SO4}$ reacts with $\pu{1 mol}$ of $\ce{NaOH}$?
Will $\pu{1 mol}$ each of $\ce{NaHSO4}$ and water be formed?
Or will $\pu{0.5 mol}$ of $\ce{H2SO4}$ will ...
0
votes
1
answer
328
views
Adjustment of calcium, magnesium and alkalinity in drinking water?
It's been a long time since I did any molar chemistry in school and despite doing a lot of reading on the subject this week, I am making a mess of some basic calculations. I hope that someone who ...
0
votes
0
answers
1k
views
Trichloroacetic acid aqueous solution
I have bought some anhydride of trichloroacetic acid (TCA) and would like to prepare an acidic solution. My problem is, that I don't know how much "powder" to mix with how much water to produce an x% ...
1
vote
2
answers
6k
views
Buffer Solution with Strong Acid?
I was thinking: Could we could make a strong acid-strong base buffer solution? Take, for instance, a mixture of 1 L $\ce{HCl}$ (0.1 M) and 1 L of aqueous salt, $\ce{Na_2SO_4}$ (also 0.1 M), note that $...
3
votes
2
answers
5k
views
Why doesn't a buffer solution change ph(Appreciably?)
Consider a buffer solution containing weak acid $\ce{CH3COOH}$ and its salt $\ce{CH3COONa}$:
\begin{align}
\ce{CH3COOH &<=> CH3COO- + H+} \tag{1}\\
\ce{CH3COONa &-> CH3COO- + Na+} \...
-2
votes
2
answers
87
views
What do we call ions without basic and acidic properties
According to Brønsted theory,
Acids are substances (molecules and ions) donating $\ce{H+}$
Bases are substances (molecules and ions) receiving $\ce{H+}$
I've been trying to find a complete Brønsted-...
-1
votes
1
answer
393
views
Why is the strong acid written in a net ionic equation? [closed]
Strong acids dissolve and break apart into ions, so in a net ionic equation, why are they written?
1
vote
1
answer
747
views
How to filter out magnesium acetate (a soluble salt) quickly and cheaply from water
I am conducting an experiment that involves reacting a weak acid and a weak base to form water and a salt with this overall equation
$$\ce{Mg(OH)2(s) + 2CH3COOH(l) <=> 2H2O(l) + Mg(CH3COO)2(aq)}...
13
votes
2
answers
10k
views
Should bromine water be called a solution?
Bromine water is a reagent which is used to test for unsaturation in organic compound. It is $2.8~\%$ bromine in water. In many places, it is refer to as bromine solution.
But it is observed that ...