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Chapter 8 Activity Goodbye Freshman Chemistry Hello Real World.

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Presentation on theme: "Chapter 8 Activity Goodbye Freshman Chemistry Hello Real World."— Presentation transcript:

1 Chapter 8 Activity Goodbye Freshman Chemistry Hello Real World

2 Activity Equilibrium up to this point has been based on a very simple model. Basically that ions (a species with a high density of charge) do not interact other than the interaction of interest.

3 Activity Real Solutions are rather complex in their nature. Ion interact with all ions in solution and other polar species. Ions will interact with water since water is a polar substance. Ions will have these water molecules in their ‘waters of hydration’.

4 Just how much water? CompoundTightly bound water CH 3 CH 2 CH 3 0 CH 3 OCH 3 1 CH 3 CN3 CH 3 COO - 5 NH 3 9 NH

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7 Activity In essence this ionic atmosphere shields the ions from each other which will diminish the interaction between the ions. i.e. For a K sp interaction the compound will be more soluble.

8 Solubity KT(s) = K + (aq) + T - (aq)

9 Potassium Tartarate

10 Glucose

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12 Activity How do we quantify this ionic atmosphere. We use a term called ionic strength. It can be calculated

13 Activity Let’s do an example. What is the ionic strength of a solution that is M Na 2 SO 4 and M NaCl We will prepare a Table and ask our first question. What ions are present?

14 Activity IonConcChargeCharge^2Product Na + SO 4 2- Na + Cl - Sum 0.5*Sum

15 0.010 M Na 2 SO 4 and M NaCl What are the concentrations?

16 Activity IonConcChargeCharge^2Product Na SO Na Cl Sum 0.5*Sum

17 0.010 M Na 2 SO 4 and M NaCl What are the charges and charges squared?

18 Activity IonConcChargeCharge^2Product Na SO Na Cl Sum 0.5*Sum

19 0.010 M Na 2 SO 4 and M NaCl Let’s Finish the Math

20 Activity IonConcChargeCharge^2Product Na SO Na Cl Sum *Sum0.08 M

21 Activity It gets even more complex. Not all soluble salts will dissociate into the individual species. We get another form that will exist in solution. This is called and Ion Pair MgSO 4 (s) + H 2 O = Mg 2+ (aq) + SO 4 2- (aq) + MgSO 4 (aq) The MgSO 4 (aq) is the ion pair.

22 Activity You can see from Appendix J in the book the equilibrium constants for this pair formation. Mg 2+ (aq) + SO 4 2- (aq) = MgSO 4 (aq) Log K = 2.23 Which means that about ~93% will be in the ion pair form if the formal concentration of MgSO 4 is 1M.

23 Activity Ion Pairing has important implications in areas such as drug transport and delivery. Any drug species must be in the proper form to give its activity.

24 Activity So how do we find the “Activity of an ion in solution”. Where A is the activity of species C  is “activity coefficient” [C] is the concentration in moles per liter

25 Activity Let’s revisit our equilibrium expression. aA + bB = cC + dD Which we write

26 Activity Lets revisit an equilibrium we have seen several times. The solubility of silver sulfide.

27 Activity How do we find the value for the “activity coefficient” ? For ionic strengths from 0.1M and down we can look up the value on a Table or we can calculate from the Extended Debye-Huckel equation.

28 Activity Where z is the ion charge,  is the ionic strength and  is the hydrated radius

29 Hydrated Radius (  )

30 Activity You could also look up the value of the activity coefficient from Table 8-1 in the text.

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33 Activity What if you want an intermediate value between two listed ionic strengths. You will need to do a linear interpolation. (Do you remember this from using log tables in high school?)

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35 Activity Effect of Ionic Strength, Ion Charge and Ion Size on Activity Coefficient (0 to 0.1M)  increases  decreases z increases the faster the departure from  from unity smaller  then the greater activity effects

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37 Activity Neutral Molecules Not charged so no ionic interaction so we assume  is 1 Gases Not charged and again we can assume that  is 1. A = P (bar)

38 Activity (In concentrated Solutions)

39 Activity What is the pH of very, very pure water? K w = A H A OH = 1.0 x A H = 1.0 x Thus pH = 7.00 What is the pH of 0.10 M NaCl? Would it still be 7.00?

40 Activity K w =  H [H]  OH [OH] Look up the  for H + and OH - on Table 8-1.  H = 0.83 and  OH = 0.76 [H + ] = [OH - ] So Solving for [H + ] = 1.26 x pH = - log A H = -log (0.83)(1.26 x ) = 6.98


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