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Titration Problem Introduction
Titrations involve reacting an acid and a base together to determine the concentration of the unknown (can be the acid or base).
Wize Tip
On a test, you could be expected to know what the equivalence point of a titration is.
Equivalence point: is the point in a titration when enough titrant has been added into the reaction to completely react with the analyte.
In other words, the acid and base completely react together so that there is no more OH- or H+ as it would have all reacted.
Example:
We may be given the concentration and volume of the base and only the concentration of the acid and asked what volume of acid was needed to reach the equivalence point?
We will need to use the equation: n=cv (n= moles, c=concentration, v=volume in L)
We could solve for the moles of the base (n=cv) and then since we are at equivalence point, we can figure out the moles of the acid.
Wize Tip
At this step, it is very important to write out a balanced chemical reaction with the acid and base reacting.
We must pay attention to the coefficients in the reaction as they are important for determining moles.
Then we could solve for the volume of the acid (use n=cv, rearrange for v: v=n/c)
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Example: Titration Problem
A 100 mL solution of HBr was titrated with 0.20 M . It was found that 38 mL of the solution was required to reach the equivalence point. What was the concentration of the solution?
1) I'll start by writing out the variables that I know:
HBr:
V=100mL=0.1L
c?
Ca(OH)2
c=0.2M
V=38mL=0.038L
2) We can find the moles of Ca(OH)2 using n=cv
n=cv
n=(0.2)(0.038)
n=0.0076 moles
3) Now that we have moles of Ca(OH)2, we can figure out moles of HBr based on the balanced chemical equation
Ca(OH)2 + 2HBr → CaBr2 + 2H2O
n=0.0076 moles for Ca(OH)2 x2 for moles of HBr
n=0.0152 moles HBr
We must take the coefficients from the balanced chemical equation into consideration!
Now when 0.0076 moles of Ca(OH)2 reacts with 0.0152 moles of HBr, there will be no more reactants left! This is exactly what we want at equivalence point!
4) Now we can solve for the concentration of HBr
n=cv
c=n/v
c=0.0152/0.1
c=0.152M