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Chapter 8 Activity

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**Homework Chapter 8 - Activity**

8.2, 8.3, 8.6, 8.9, 8.10, 8.12

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**Question 8.2 Q: Which statements are true?**

In the ionic strength, m, range of M, activity coefficients decrease with: a) increasing ionic strength b) increasing ionic charge c) decreasing hydrated radius All are true!!

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**Question 8.3 Calculate the ionic strength of**

M KOH M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0087M(+1) M(-1)2] = M Remember for +1/-1 systems: Ionic strength, m = Molarity, M

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**Question 8.3 (cont’d) Calculate the ionic strength of**

M KOH M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0002M(+3) M(-1)2] = M

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Question 8.6 Calculate the activity coefficient of Zn2+ when m = M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. a) =-0.375 g=0.422

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Question 8.6 (cont’d) Calculate the activity coefficient of Zn2+ when m = M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. = 0.432

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Question 8-9 Calculate the concentration of Hg22+ in saturated solutions of Hg2Br2 in M KNO3. Hg2Br2(s) D Hg Br- Ksp=5.6x10-23 some -x +x +2x some-x +x +2x I C E

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8-10. Find the concentration of Ba2+ in a M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 Ksp = 7.11 x 10-11 I C E some -x +x +2x some-x +x +2x

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8-10. Find the concentration of Ba2+ in a M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 Ksp = 1.5 x 10-9 I C E some -x +x +2x some-x +x 0.1+2x

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8-10. Find the concentration of Ba2+ in a M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 Ksp = 1.5 x 10-9 I C E some -x +x +2x some-x +x 0.1+2x X = 6.57 x 10-7

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Question 8-12 Using activities correctly, calculate the pH of a solution containing M NaOH plus M LiNO3. What is the pH if you neglected activities? (m) = ½ (c1z12+ c2z22 + …) = ½ [0.010M(+1) M(-1) M(+1) M(-1)2] = M gOH = pH = AH = [H+]gH

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Question 8-12 (cont’d) Using activities correctly, calculate the pH of a solution containing M NaOH plus M LiNO3. What is the pH if you neglected activities? pH = 11.94

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Question 8-12 (cont’d) Using activities correctly, calculate the pH of a solution containing M NaOH plus M LiNO3. What is the pH if you neglected activities? pH ~ -log[H+] =

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Finally Calculate the pH of a solution that contains 0.1 M Acid and 0.01 M conjugate base Calculate the pH of a solution containing 0.1 M Acid and 0.05 M conjugate base. Calculate the pH of a solution containing 0.1 M Acid and 0.1 M conjugate base.

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Acid/Base Titrations

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**Titrations Titration Curve – always calculate equivalent point first**

Strong Acid/Strong Base Regions that require “different” calculations B/F any base is added Half-way point region At the equivalence point After the equivalence point

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**Strong Acid/Strong Base**

50 mL of M KOH Titrated with M HBr First -find Volume at equivalence M1V1 = M2V2 (0.050 L)( M) = V V = 10.0 mL

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**Strong Acid/Strong Base**

50.00 mL of M KOH Titrated with M HBr Second – find initial pH pH = - logAH ~ -log [H+] pOH = -logAOH ~ -log [OH-] pH = 12.30

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**Strong Acid/Strong Base**

50 mL of M KOH Titrated with M HBr Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) (~6 ml) Before After mol mol mol 0 mol Limiting Reactant mol mol pH = 11.8

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**Strong Acid/Strong Base**

50 mL of M KOH Titrated with M HBr Fourth – find pH at equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) Before After mol mol 0 mol 0 mol mol mol pH = 7.0

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**Strong Acid/Strong Base**

50 mL of M KOH Titrated with M HBr Finally – find pH after equivalence point 12 ml KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) Limiting Reactant Before After mol mol 0 mol mol mol pH = 2.5

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**Titration of WEAK acid with a strong base**

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**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH First, calculate the volume at the equivalence-point M1V1 = M2V2 ( L) M = M (V2) V2 = L or 25.0 mL

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**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH Second, Calculate the initial pH of the acetic acid solution

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**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH Third, Calculate the pH at some intermediate volume

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**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH Fourth, Calculate the pH at equivalence

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**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH Finally calculate the pH after the addition mL of NaOH

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