1. Challenges are what make life interesting; overcoming them is what makes life meaningful. -Joshua J. Marine-

2. TITRATION – A METHOD TO DETERMINE THE AMOUNT OF AN ACID OR A BASE IN SOLUTION. A SOLUTION OF KNOWN CONCENTRATION IS ADDED FROM A BURET TO A MEASURED VOLUME OF A SOLUTION OF UNKNOWN CONCENTRATION. THE STOICHIOMETRIC POINT IS DETERMINED THROUGH THE USE OF AN INDICATOR OR pH METER. A PLOT OF pH VERSUS VOLUME OF TITRANT IS CALLED A TITRATION CURVE.

3. LET’S ASSUME WE ARE TITRATING A STRONG ACID WITH A STRONG BASE. H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  H 2 O (l) + Na + (aq) + Cl - (aq) Or, getting rid of the spectator ions: H + (aq) + OH - (aq)  H 2 O (l) Na + AND Cl - DON’T ACT AS ACIDS OR BASES, SO THE pH AT THE STOICHIOMETRIC POINT OR EQUIVALENCE POINT WOULD BE 7. THE STOICHIOMETRIC POINT OR EQUIVALENCE POINT IS THE POINT AT WHICH YOU HAVE ADDED JUST ENOUGH STRONG BASE TO EXACTLY REACT WITH ALL OF THE STRONG ACID. #MOLES ACID = #MOLES BASE V a X M a = V b x M b

4. WE COULD INTRODUCE A NEW UNIT, MILLIMOLES 1 MILLIMOLE = 1 mm = MOLE/1000 SO, IN THE EQUATION V a x M a = V b x M b if the volume is in ml # MILLIMOLES ACID = #MILLIMOLES BASE MOLARITY = #MOLES/#LITERS = #MILLIMOLES/#MILLILITERS

5. LET’S ASSUME WE ARE TITRATING 50 ML OF 0.200 M HNO 3 WITH 0.1 M NaOH. AT THE START, WE HAVE H +, NO 3 -, H 2 O [H + ] = 0.200 M pH = -log (0.200) = 0.699 IF WE ADD 10 ml 0.1 M NaOH 10 ml x 0.1 M = 1.0 mmole OH - WE STARTED WITH 50 x 0.2 = 10 mmole H + AMOUNT OF H + REMAINING = 10-1 = 9 mmole NEW VOLUME = 50 + 10 = 60 ml [H + ] = 9 mmole/60 ml = 0.15 M pH = - log (0.15) = 0.82

6. IF WE ADD 20 ml (TOTAL) 0.1 M NaOH # mmole H + REMAINING = 10 – 2 = 8 NEW VOLUME = 70 ml [H + ] = 8 mmole/70 ml = 0.114 M pH = - log (0.114) = 0.943 IF WE ADD 30 ml (TOTAL) 0.1 M NaOH # mmole H + REMAINING = 10 – 3 = 7 NEW VOLUME = 80 ml [H + ] = 7 mmole/80 = 0.0875 M pH = -log (8.75 x 10 -2 ) = 1.06

7. IF WE ADD 40 ml 0.1 M NaOH #mmole H + REMAINING = 10 – 4 = 6 NEW VOLUME = 90 ml [H + ] = 6 mmole/90 ml = 0.067 pH = - log (0.067) = 1.17 IF WE ADD 50 ml 0.1 M NaOH #mmole H + REMAINING = 10 – 5 = 5 NEW VOLUME = 100 ml [H + ] = 5 mmole/100 = 0.05 M pH = -log (0.05) = 1.30

8. IF WE ADD 80 ML 0.1 M NaOH # mmole H + REMAINING = 10 – 8 = 2 NEW VOLUME = 130 ml [H + ] = 2 mmole/130 ml = 0.0154 M pH = -log (0.0154) = 1.81 IF WE ADD 100 ml 0.1 M NaOH # mmole H + REMAINING = 10 – 10 = 0 MAJOR SPECIES IN SOLUTION Na +, Cl -, H 2 O Na + and Cl - ARE NEITHER ACIDS OR BASES, SO pH = 7

9. IF WE ADD 120 ml 0.1 M NaOH # mmole OH - = 20 x 0.1 = 2 mmole NEW VOLUME = 170 ml [OH - ] = 2 mmole/170 ml = 1.18 x 10 -2 pOH = - log (1.18 x 10 -2 ) = 1.93 pH = 14 – 1.93 = 12.1 IF WE PLOT pH VS VOLUME, WE GET THE TITRATION CURVE, AND IT WOULD LOOK LIKE:

10. SO, IF YOU HAVE SOME WAY OF DETERMINING WHEN THE EQUIVALENCE POINT (STOICHIOMETRIC POINT) OCCURS, YOU HAVE A WAY OF DETERMINIG THE CONCENTRATION OF AN UNKNOWN ACID OF BASE SOLUTION. WE CAN DO THIS USING A pH METER, WHICH USES AND ELECTRODE SENSITIVE TO HYDROGEN ION CONCENTRATION.

11. YOU COULD ALSO USE A pH INDICATOR DYE. THESE ARE ORGANIC DYES WHOSE COLORS DEPEND ON pH. THEY CHANGE COLOR OVER A SHORT pH RANGE. UNIVERSAL INDICATOR