Acids and Bases

Buffers

A buffer system is a solution that can absorb moderate amounts of acid or base without a significant change in pH.  They contain relatively high concentrations of a weak acid and a salt of that acid – or a weak base and it’s salt.  Buffers control pH because they contain an acid that can react with added base and also react with any added acid.

Buffers Example: the human bloodstream, where blood is buffered principally by the hydrogen carbonate ion: HCO H 3 O +1 → H 2 CO 3 + H 2 O HCO OH -1 → H 2 O + CO 3 -2 This buffer system can neutralize the moles of carbonic acid, plus phosphoric acid and lactic acid that the body produces daily.

Titration Author: J.A.Freyre

Titration TitrationTitration –Analytical method (laboratory) in which a standard solution is used to determine the concentration of an unknown solution. standard solution unknown solution Courtesy Christy Johannesson Standard solution~ Any solution for which the concentration is precisely known.

Equivalence point (endpoint)Equivalence point (endpoint) –Point at which equal amounts of H 3 O + and OH - have been added. –Determined by… dramatic change in pH Titration indicator color change Courtesy Christy Johannesson

Acid-Base Titration

14.55 mL

23 24 How to read a buret volume mL (not mL) mL?

indicator-changes color to indicate pH change e.g. phenolpthalein is colorless in acid and pink in basic solution Pirate…”Walk the plank” once in water, shark eats and water changes to pink color pH endpoint equivalence point base 7 pink Titration

equivalence point pH Volume of M NaOH added (mL) Titration of a Strong Acid With a Strong Base Solution of NaOH Solution of NaOH Solution of HCl H+H+ H+H+ H+H+ H+H+ Cl Cl - Na + OH - Acid-Base Titrations Adding NaOH from the buret to hydrochloric acid in the flask, a strong acid. In the beginning the pH increases very slowly. Adding additional NaOH is added. pH rises as the equivalence point is approached. Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point.

equivalence point pH Volume of M NaOH added (mL) Titration of a Strong Acid With a Strong Base NaOH added (mL) pH Titration Data Solution of NaOH Solution of NaOH Solution of HCl H+H+ H+H+ H+H+ H+H+ Cl - Na + OH - 25 mL phenolphthalein - colorless phenolphthalein - pink Bromthymol blue is best indicator: pH change Yellow Blue

Titration of a Strong Acid With a Strong Base equivalence point pH Volume of M NaOH added (mL) Color change methyl violet Color change bromphenol blue Color change bromthymol blue Color change phenolpthalein Color change alizarin yellow R (20.00 mL of M HCl by M NaOH) Hill, Petrucci, General Chemistry An Integrated Approach 2nd Edition, page 680

equivalence point pH Volume of M NaOH added (mL) Titration of a Weak Acid With a Strong Base NaOH added (mL) pH Titration Data Titration of a Weak Acid With a Strong Base Phenolphthalein is best indicator: pH change

equivalence point pH Volume of M HCl added (mL) Titration of a Weak Base With a Strong Acid HCl added (mL) pH Titration Data Titration of a Weak Base With a Strong Acid 50.0

Titration Concentration of the unknown can be determined mathematically if you know: –The balanced neutralization reaction –The concentration of the standard –The volume of the standard used –The volume of the unknown used

Titration ? M of HCl 30.0 mL of 2.0 M of NaOH If it requires 10.5 mL of ? M HCl to titrate 30.0 mL of 2.0 M NaOH to its endpoint: what is the concentration of the HCl? M 1 V 1 = M 2 V 2 M V = M V H+H+ H+H+ OH - HCl (aq)  H + (aq) + Cl - (aq) 0.1 M H 2 SO 4 (aq)  2 H + (aq) + SO 4 2- (aq) 0.1 M “0.2 M”0.1 M proper term is Normality (N) M V n = M V n H+H+ H+H+ OH - Al(OH) 3 (aq)  Al 3+ (aq) + 3 OH - (aq) 10.5 mL HCl must be ~ __x more concentrated than the NaOH. 6 (x M)(10.5 mL) = (2.0 M)(30.0 mL) X = 5.7 M 30.0 mL of NaOH with bromthymol blue indicator muriatic acid sunnyside 0.1 molar H 2 SO 4 is 0.2 normal 10.5 mL of HCl Endpoint of titration is reached…color change.

Titration moles H 3 O + = moles OH - M  V  n = M  V  n M:Molarity V:volume n:# of H + ions in the acid or OH - ions in the base Courtesy Christy Johannesson

Solution of NaOH Solution of KOH Solution of H 2 SO mL Titration 42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H 2 SO 4 Find the molarity of H 2 SO 4. H3O+H3O+ M = ? V = 50.0 mL n = 2 OH - M = 1.3M V = 42.5 mL n = 1 MV# = MV# M(50.0mL)(2) =(1.3M)(42.5mL)(1) M = 0.55M H 2 SO 4 Courtesy Christy Johannesson 2 KOH + H 2 SO 4 → K 2 SO H 2 O