19
HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) NO EXTRA CREDIT FOR CHAPTER 18 HW Lab Wednesday/Thursday – Locker check-out and questions about the final

HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Embed Size (px)

Citation preview

Page 1: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all)

NO EXTRA CREDIT FOR CHAPTER 18 HW

Lab Wednesday/Thursday – Locker check-out and questions about the final

Page 2: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Sig., Figs., and Logs When you take the log of a number written in scientific

notation, the digits before the decimal point come from the exponent, and the digits after the decimal point come from the decimal part of the number.

Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig. figs. are the digits after the decimal point in the log.

6log(2.0 10 ) 62.0l g( )10o 6= log(10 )6= log( ) + log10 (2.0)

= 66 = + 0.30 30300.303= 030 6 6 + = . 303030

6log( )2.0 = 010 6.3

Page 3: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

pKWhat does “p” mean?

p___ = -log___ (pH = - log H+, pOH = -logOH-)A way of expressing the strength of an acid or base

is pK.pKa = −log(Ka), Ka = 10−pKa

pKb = −log(Kb), Kb = 10−pKb

The stronger the acid, the smaller the pKa.Larger Ka = smaller pKa

The stronger the base, the smaller the pKb.Larger Kb = smaller pKb

Page 4: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Finding pH of a Strong Acid or Strong Base Solution

For a monoprotic strong acid [H+] = [HA].

For 0.10 M HCl, [H+] = 0.10 M and pH = 1.00

For 0.025 M HNO3, [H+] = 0.025 M and pH = 1.60

For a strong ionic base, [OH−] = (number OH−) × [Base].

For 0.15 M Ca(OH)2, [OH−] = 0.30 M and pH = 13.48

For 0.15 M KOH, [OH−] = 0.15 M and pH = 13.18

Page 5: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

pH of an Acid/Base Mixture

17.20 mL of a 0.09000 M barium hydroxide solution is added to 25.00 mL of 0.1234 M hydrochloric acid.1) What are the pH and pOH of the original barium hydroxide?2) What is the pH of the original hydrochloric acid solution?3) What is the pH of the solution after the two solutions are mixed?

pOH Ba(OH)2(aq) = – log(0.18000) = 0.74473

pH Ba(OH)2(aq) = 14 – 0.74473 = 13.25527

pH HCl(aq) = – log(0.1234) = 0.9087

Page 6: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

pH of an Acid/Base Mixture

17.20 mL of a 0.09000 M barium hydroxide solution is added to 25.00 mL of 0.1234 M hydrochloric acid.1) What are the pH and pOH of the original barium hydroxide?2) What is the pH of the original hydrochloric acid solution?3) What is the pH of the solution after the two solutions are mixed?

+(aq)

0.1234 1mol H 0.02500 0.003085

1 1

+(aq) (aq) +

(aq) (aq)(aq) (aq)

HCl H HCl H

HCl HCl

mol molL mol

L mol

(aq)

0.09000 2mol OH 0.01720 0.003096

1 1

2(aq) (aq)

2(aq) (aq)2(aq) 2(aq)

Ba(OH) OH Ba(OH) OH

Ba(OH) Ba(OH)

mol molL mol

L mol

0.003096 0.003085 0.000011 +(aq) (aq) (aq) OH H OH excess mol mol mol

40.000011 2.607 10

0.02500 0.01720

(aq) OH excess

OH L L

molM

pOH = – log(2.607x10–4) = 3.5839

pH = 14 – 3.583920 = 10.4161

Page 7: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Strong Bases

• The stronger the base, the more willing it is to accept H+.

• For ionic bases, practically all units are dissociated into OH– or accept H+.Strong electrolyteMulti-OH strong bases completely

dissociated

Page 8: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Weak Bases

• In weak bases, only a small fraction of molecules accept H+.Weak electrolyteMost of the weak base molecules do

not take H from water.Much less than 1% ionization

in water

• [HO–] << [weak base]• Finding the pH of a weak base

solution is similar to finding the pH of a weak acid.

Page 9: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

TitrationIn an acid–base titration, a solution of known

concentration from a burette (titrant) is slowly added to a solution of unknown concentration (analyte) until the reaction is complete. When the reaction is complete we have reached the endpoint of the

titration.An indicator may be added to determine the

endpoint.When the moles of H+ = moles of OH−, the titration

has reached its equivalence point.

Page 10: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration

Page 11: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration Curve A plot of pH versus the amount of added titrant. Prior to the equivalence point, the known solution in the

flask is in excess, so the pH is closest to its pH.

Page 12: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration Curve A plot of pH versus the amount of added titrant. The inflection point of the curve is the equivalence

point of the titration.

Page 13: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration Curve A plot of pH versus the amount of added titrant. The inflection point of the curve is the equivalence

point of the titration. The pH of the equivalence point depends on the pH of

the salt solution. Equivalence point of neutral salt, pH = 7 Equivalence point of acidic salt, pH < 7 Equivalence point of basic salt, pH > 7

Page 14: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration Curve A plot of pH versus the amount of added titrant. Beyond the equivalence point, the unknown solution in

the burette is in excess, so the pH approaches its pH.

Page 15: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in

differences in the titration curve at the equivalence point and excess acid region.

Page 16: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in

differences in the titration curve at the equivalence point and excess acid region.

The initial pH is determined using the Ka of the weak acid.

The pH in the excess acid region is determined as you would determine the pH of a buffer.

The pH at the equivalence point is determined using the Kb of the conjugate base of the weak acid.

The pH after equivalence is dominated by the excess strong base. The basicity from the conjugate base anion is negligible.

Page 17: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration of a Polyprotic AcidIf Ka1 >> Ka2, there will be two equivalence points

in the titration.

Titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH

Page 18: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration of a Polyprotic AcidIf Ka1 >> Ka2, there will be two equivalence points

in the titration.

Page 19: HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all),

Titration of a Polyprotic AcidIf Ka1 >> Ka2, there will be two equivalence points

in the titration.The closer the Ka’s are to each other, the less

distinguishable the equivalence points are.