Arrhenius's definition of an acid is "a substance that produces H+ ions when dissolved in water". His definition of a base is "a substance that produces OH- ions when dissolved in water. These definitions are sound but we would like to have a definition which is more general. Bronsted and Lowry defined acids and bases as: "an acid is a proton donor" and "a base is a proton acceptor". From the B-L definitions we see that ammonia (NH3) is a base as it will accept a proton to form NH4+. HCl is an acid as it will donate a proton (HCl --> H+ + Cl-). We will study these concepts in more detail in a later section. For now we are discussing reactions so let's look at reactions of acids and bases.
In high school you learned that a strong acid and a strong base will produce a salt and water:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
To a scientist the word "salt" means any ionic compound whose cation comes from a base and whose anion comes from an acid. Since the above compounds are strong electrolytes, we know that they completely dissociate in water so what is the ionic equation for the above reaction?
HCl(aq) + NaOH(aq) H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) and then
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) NaCl(aq) Na+(aq) + Cl-(aq)
So let's cancel the spectator ions and see what we have:
H+(aq) + OH-(aq) H2O(l)
This is the net ionic reaction for any reaction of a strong acid with a strong base. Such reactions are called neutralization reactions because the solution is neither acid nor base upon completion of the reaction. Below is another example:
HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)
What about reactions of a weak acid and a strong base? We have seen that acetic acid (HC2H3O2) is a weak acid and we know that NaOH is a strong base so let's see what happens when we mix them in solution. By the way, for beginning chemistry we write acetic acid as the above with one H atom in front of the C atom to indicate that that H atom is the acidic hydrogen atom (the one which will be donated -- BL theory). Since acetic acid is a weak electrolyte it will only be very weakly dissociated in solution. But NaOH will be completely dissociated so that we will have:
HC2H3O2 + Na+(aq) + OH-(aq) ??????
This solution does not sit around long with nothing happening because the hydroxide ion is such a strong base (proton acceptor) that it will remove the acidic hydrogen from acetic acid:
OH-(aq) + HC2H3O2 H2O(l) + C2H3O2-(aq)
And now the Na+ ions in solution are attracted to the acetate ion so that we have
C2H3O2-(aq) + Na+(aq) NaC2H3O2(aq)
The overall reaction is then
HC2H3O2(aq) + NaOH(aq) NaC2H3O2(aq) + H2O(l)
where the salt formed is called sodium acetate.
Another important point is that the H+ ion produced by an acid in solution is rapidly hydrated to produce what is called the hydronium ion:
Many texts will use the notation H+ to represent the hydrated proton rather than specifically writing it out as the above.
So let's work a stoichiometric problem with acids and bases. What volume of a 0.200 M KOH solution is needed to neutralize 50 ml of a 0.600 M solution of HNO3?
I would work this by first writing out the chemical reaction which will occur:
KOH + HNO3 H2O + KNO3
I see that the stoichiometric coefficients are all one so I know that a mol of KOH reacts with one mole of nitric acid. I know that 50 mls of 0.600 M nitric acid will give the following number of moles of nitric acid:
0.600 mol/liter * (0.050 liter) = 0.030 mol nitric acid = 0.030 mol H+ = 0.030 mol NO3-
Since all stoichiometric coefficients are one, we will need a one to one ratio of K+ ions to NO3- ions. Therefore I will need 0.030 moles of KOH. So what volume of KOH will I need to supply 0.030 moles of KOH?
0.200 mol/liter * (X liter) = 0.030 mol
Solving for X gives 0.150 liters. Therefore I will need 150 ml of 0.200 M KOH to completely neutralize the 50 ml of 0.600 M nitric acid. Easy!!
Titrations of acids and bases is just the use of an accurate delivery system of an acid (or a base) solution to add to a solution of a base (or acid). A buret can be use to deliver the solution accurately into a known volume of the other solution. When we have titrated the correct amount of acid (or base) into the base (or acid) solution so that neutralization has occurred, we are said to have reached the endpoint or the equivalence point or the stoichiometric point. In the lower level courses this endpoint is usually determined by using an indicator which will change color when the endpoint has been reached. Actually the endpoint is not necessarily exactly equal to the equivalence point, but that is another story for another day.
The above could have been done as an acid-base titration by stating the problem as: 50 ml of a 0.600 M solution of nitric acid was titrated into a 150 ml solution of KOH when the indicator changed color. What was the concentration of the KOH solution?
You see that the solution is the same as the above except that the final working equation is:
X mol/liter * 0.150 liter = 0.030 mol
X = 0.200 M
Now what do you have to do? Practice, practice, practice.
|Now take a practice quiz to help you understand if you understand the basic concepts.|
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Web Author: Dr. Leon L. Combs
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