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Published byRosamond McDaniel Modified over 9 years ago
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THE MATHEMATICS IN A TITRATION CURVE (WITH A LITTLE BASE 10 AND LOGARITHM ARITHMATIC ADDED)
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INTRODUCTION OF BASIC CONCEPT OF TITRATION CURVES USING TITRATION OF STRONG ACID WITH STRONG BASE
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How do calculations of pH where the solution changes from acidic to basic compare to the curve of the titration graph?
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Recall: What is a mole? 6.022 x 10 23 molecules, atoms, or ions # M represents the fraction ( # moles)/(1 liter of solution) and this value tells us the concentration of the solution pH = -- log [H + ]
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An example of a strong acid base reaction is: Add drops of NaOH To a solution of HCL --- Add an indicator which changes color with pH (phenalthalien turns from clear to pink)
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Background Concepts: Strong Acid titrated with a Strong Base Consider the titration of a 50ml sample of 1M HCl titrated with 1M NaOH Add drops of NaOH To 50 ml of 1M solution of HCL
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3 important Calculations of pH when the volume of NaOH added is close to Equivalency Point:
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ml of NaOH = 49.99 ml or 0.04999 liters total volume in beaker is approximately 100 ml There is 0.01 ml of H + ions which have not paired with OH - to form water. [H + ] = 10 ^ -4 pH = 4
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When the ml of NaOH = 50 ml then total volume of added NaOH equals volume of HCl all H + ions have combined with OH – ions to form water neutral pH = 7
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note: K w = = [H + ] [OH - ] for this volume of NaOH ---- [H + ] =[OH - ] giving the equation K w =[H + ] 2 and [H + ] = 10^-7. This reviews laws of exponents – taking square root means dividing exponent by 2.
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When ml of NaOH added = 50.01 ml There is 0.01 ml of excess OH — ions in solution. [OH - ]= 10^-4 Since [H + ] [OH - ] = 10^-14 then [H + ] = 10^-14 / [OH - ] [H + ] = 10^-14/10^-4 or [H + ]= 10^-10 and pH = 10
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CONCLUSION: The pH changes 6 points when the volume of NaOH changes 2 ml close to the equivalency point for this mixture of strong acid and strong base.
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Relate this to the titration curve for strong acids and strong bases.
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