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The classic example of a titration problem in chemistry is the titration of a strong acid with a strong base. With these types of titrations, the equivalence point (the point at which the moles of the acid and base are equal and thus cancel each other out chemically) is clearly defined through the use of an indicator. The acid and base are usually contained as solutions of particular concentrations, and the volumes of both species utilized are monitored.
In general, there are four different numerical values that are necessary in order to do a titration problem. Three of these values are usually given, and the purpose of the problem is to calculate the fourth value. The concentrations of the acid and base solutions and the volumes of both solutions are the four numerical values in question. When the concentration of a solution is multiplied by the volume of solution used, the number of moles of the chemical species is obtained. Always remember that at the equivalence point, the number of moles of acid is equal to the number of moles of base. So whatever species (the acid or base) you are given both the concentration and volume of, multiply those values together to get the number of moles. Then that number of moles can be used to calculate either the concentration or the volume used of the other chemical species.
There are various titrations, but you should only be required to know a strong/strong titration for high school chemistry. When acids and bases are mixed, water and a neutral salt are formed.
HA + BOH -> H2O + BA
Strong acids and bases are strong electrolytes, so they dissociate completely. This means that the molarity of the chemical is the same as the molarity of the individual ions.
For example, 1.0 M of HCl dissociates into 1.0 M of H+ and 1.0 M Cl-.
With a strong base and strong acid, the pH is 7 when there are equal moles of H+ and OH-. You can simply set up an equation:
[acid] x volume = [base] x volume
moles of acid = moles of base
To find out the molarity or volume needed to neutralize an acid (or vice versa), you will need to plug in the information you are given into the equation above.
For example, to neutralize a 75 mL 0.5 M solution of HCl, you need x amount of a 1.0 M solution of NaOh.
(0.5 M)(75 mL) = (1.0)x
After solving for x, you will found out that you need 37.5 mL of NaOH.
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