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Acid-Base calculations

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Presentation on theme: "Acid-Base calculations"— Presentation transcript:

1 Acid-Base calculations
Molarity calculations play an important part in acid-base reaction stoichiometry Much of what we will learned in Chapter 3 will be used here.

2 Molarity M = mol/L or M = mmol/mL
we can use moles and liters, or millimoles and milliliters, and the molarity is still the same.

3 Similarities between acid-base and other reaction calculations
We still compare moles to moles, not volumes to volumes or molarities to molarities. Additionally, knowing the limiting reactant is very important (i.e, what will run out first—acid or base?)

4 Some examples what volume of M NH3 is required to neutralize 22.0 mL of 12.0 M HCl? 25.0 mL of M Ca(OH)2 added to 10.0 mL of HNO3. Is the solution now acidic or basic? how many moles excess acid or base are in the solution? how much additional Ca(OH)2 or HNO3 sol’n required to neutralize solution?

5 TITRATIONS Combining a known concentration with an unknown concentration solution. Titrant: The solution of one reactant (usually of unknown concentration) that is carefully added to the solution of the other reactant until the resulting solution is just neutralized (no excess acid or base). How do we know when to stop?

6 Titrations (con’t) indicators: How to measure the volume of titrant?
Buret: equivalence point: The point where _____________ _______________ amounts of acid and base have reacted. end point: The point where the indicator ____________ ___________. For accurate work, one wants the end point and equivalence point to coincide with each other.

7 Primary and secondary standards
Reading on standardization: your text goes over the requirements of a primary standard. You should be familiar with these requirements. Primary standards are used to determine the concentration of solutions, which become secondary standards. Example: KHP and NaOH.

8 EQUIVALENT WEIGHTS AND NORMALITY
One mole of acid is _____________________________ But, one equivalent of acid contains ______________________________. The equivalent weight, then, corresponds to mass/equivalents.

9 normality number of equivalents per liter, or N = eq/L = meq/mL
N = M  eq/mol Let’s do a couple of examples…

10 EQUIVALENTS in acid/base reactions
1 eq acid always reacts with 1 eq base. Va Na = Vb Nb


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