University:
Jackson College
Course:
CEM-141 | General Chemistry I
Academic year:

2024
Views:

254

Pages:

7
Author:

Kael Barker

Chemistry 12 UNIT 4 ACIDS AND BASES PACKAGE #4 ACIDS STRONG monoprotic CONSIDER WEAK polyprotic polyprotic H2SO4 ⇄ H+ + HSO4 - If [H+]1 >> K2 ignore 2nd dissoc. calculate in 2 stages since [H+]1 < K2 commonly monoprotic If [H+]1 < K2 do 2nd dissoc. POLYPROTIC ACIDS: SULPHURIC H2SO4 OXALIC H2C2O4 SULPHUROUS H2SO3 PHOSPHORIC H3PO4 CARBONIC H2CO3 HYDROSULPHURIC H2S ARSENIC H3AsO4 K1 = 6.0 x 10-3 K2 = 1.0 X 10-7 K3 = 3.0 X 10-12 MALONIC H2C3H2O4 K1 = 1.5 x 10-3 K2 = 3.0 X 10-6 PHTHALIC H2C8H4O4 K1 = 1.3 x 10-3 K2 = 3.9 X 10-6 MALEIC H2C4H2O4 K1 = 1.4 x 10-2 K2 = 8.6 X 10-7 K2 = 1.3 X 10-13 DISSOCIATION MECHANISM for H3PO4: K1: H3PO4 ⇄ K2: H2PO4- ⇄ K3: HPO42- ⇄ H+ + H2PO4 H+ + HPO4 2H+ + PO4 3- --------------------------------------------------------------------------------------------------------------------H3PO4 ⇄ H+ + H2PO4 Ka = K1 = 7.5 x 10-3 I R E 1.0 M -y 1.0-y -+y y -+y y K 1= H3PO4 ⇄ H+ + H2PO4 - 7.5 x 10-3 = y2 assume y<<1.0 1.0-y -3 2 7.5 x 10 = y y = 0.0866 = [H+] pH = 1.06 --------------------------------------------------------------------------------------------------------------------H3PO4 ⇄ 2H+ + HPO4 2Ka = K1K2= I 1.0 M --(7.5 x 10-3)(6.2 x 10-8) R -y +2y +y =(4.65 x 10-10) E 1.0-y 2y y K 2= H2PO4- ⇄ H+ + HPO4 2- 4.65 x 10-10 = (y)(2y)2 assume y<<1.0 1.0-y -10 4.65 x 10 = 4y3 y = 4.88 x 10-4 x 2= [H+] = 9.76 x 10-4 M pH = 3.01 --------------------------------------------------------------------------------------------------------------------H3PO4<====> 3H+ + PO4 3Ka = K1K2K3= I 1.0 M --7.5 x 10-3)(6.2 x 10-8)(2.2 x 10-13) R -y +3y +y =(1.0 x 10-22) E 1.0-y 3y y K 3= HPO4- ⇄ H+ + PO4 3- 1.0 x 10-22 = (y)(3y)3 assume y<<1.0 1.0-y 1.0 x 10-22 = 27y4 y = 1.4 x 10-6 x 3= [H+] = 4.16 x 10-6 M pH = 5.38 --------------------------------------------------------------------------------------------------------------------POLYPROTICS RELEASE HYDROGENS WITH INCREASING DIFFICULTY K2<> K2 then [H+]2 = K2 and there is no effect If [H+]1 ~ K2 or [H+]1 < K2 then there is a significant effect DO A TWO STAGE CALCULATION! GENERALLY the second / third dissociations can be ignored for weak acids unless the solution is extremely dilute (check [H+]1 versus K2 ) GENERALLY FOR H2SO4 a two stage problem must be done IMPORTANT NOTE: GENERALLY WEAK POLYPROTIC ACIDS WILL HAVE INSIGNIFICANT SECOND AND THIRD DISSOCIATIONS, UNLESS THEY ARE EXTREMELY DILUTED. eg. 0.0000010 M H3PO4 has [H+] = 10-5 M Therefore assumptions such as y<< 0.0000010 M are no longer valid and therefore you must calculate the second and if applicable third dissociations. INDICATORS Indicators are substances that must be added to a flask when doing a titration to bring about a colour change at the endpoint of the titration. (i.e. when all of the H3O+ is neutralized by OH- or vice versa). Transition point: Equivalence point: Indicators may be weak acids or bases whose conjugate has a different characteristic colour. OR Indicators may be weak acids or bases which dissociate in water and form ionic compounds which have different colours. If Indicators are weak acids - when they lose their H+ ion to form the conjugate base, they change colour: HIn + H2O ⇄ H3O+ + Inweak acid weak base (one colour) (different colour) Ka expression for indicators: KIn = [H+][In-] [HIn] Consider the titration of an acid with a base: Inside the flask: HX + HIn + H2O ⇄ H3O+ + X- + HIn the colour does not change immediately Add OH- drop by drop from the burette The OH- will react with the strongest acid H3O+ H3O+ + OH- ⇄ 2H2O Start to use up the H3O+, the initial reaction shifts right Eventually all of the HX dissociates and all of the H3O+ is consumed by the OHWe have reached the endpoint. At the endpoint, Ka = [H+] eg. If HX has a Ka = 10-7 then its endpoint occurs when [H+] = 10-7 or pH = 7 If you choose the indicator correctly, the endpoint (transition point) = the equivalence point (stoich) NOTE: the total [H+] is from 1) the water dissociation 2) the acid dissolved 3) the indicator itself The acid dissolved is by far the greatest contribution On the next drop of OH-, the HIn now reacts (because the HX and H3O+ are totally consumed) HIn + H2O ⇄ H3O+ + InThere is a colour change as HIn changes to InConsider [HIn] > [In-] the acid colour dominates HIn + H 2O ⇄ H3O+ + In- [HIn] < [In-] the base colour dominates HIn H 2O ⇄ H 3O + [HIn] = [In-] intermediate colour occurs The indicators endpoint + + In- At endpoint: KIn = [H+][In-] [HIn] Ka = [H+] since [HIn] = [In-] Consider the titration of a base with an acid: In a base the HIn is in the form of In- because the H from the HIn has been donated to form the conjugate base InH3O+ + In+ OH- ⇄ 2H2O + Inweak strong base base As soon as all of the OH- is consumed, then on the next drop, the H3O+ reacts with the InH3O+ + In⇄ HIn + H2O one colour another colour We have reached the endpoint. UNIVERSAL INDICATOR -an indicator solution which changes colour several times over several pH ranges For example: Mixing ethyl orange (which changes from red to yellow at pH = 4.1; HEth is red) with bromthymol blue (which changes from yellow to blue at pH = 6.8; Brom- is blue) yields a universal indicator. When this mixed indicator is added to a solution of pH = 6, what colour will the indicator turn? The indicator turns yellow (it is at the basic side of the HEth indicator - the yellow side, and on the acidic side of the BBlue indicator - the yellow side) SAMPLE QUESTIONS: 1. An indicator, HIn, is red in acids and blue in bases. What is the colour of the anion, In-? Answer 2. The indicator Clayton Yellow changes from yellow to amber at pH = 12.7. What is the Ka for the inidicator? Answer 3. BLUE pKa = pH = 12.7 Ka = 10-12.7 = 2x10-13 (Ka = [H+]) Given the following observations, determine the pH of the sample: Indicator added Methyl orange Thymol blue Bromthymol blue Colour changed yellow yellow yellow Answer pH is between 4.4 and 6.0 INDICATOR USE: 1) Select an indicator which has an endpoint matching the range of the system studied eg. Bromothymol blue in strong acid / strong base titration - more to come on this later 2) Select an indicator with suitable colours -don't use phenolphtalein in blood pH study 3) Use an appropriate (small) amount Don't use so much that your test sample is changed wrt acidic vs. basic -due to dissociation of the indicator HIn + H2O ⇄ H3O+ + In- SEE BOOKLET PAGE 7 pH ranges of indicators and colour changes SEE "APPENDIX A" handout Relative strengths of acids in aq solutions at RT includes indicators and their Ka values SEE 'INDICATORS' handout See chart of pH ranges and colour changes

Acids and Bases Unit

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