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1 Example 25 mL of 0.100 M AgNO 3 are mixed with 35 mL of 0.050 M K 2 CrO 4. Find the concentration of each ion in solution at equilibrium if k sp of Ag.

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Presentation on theme: "1 Example 25 mL of 0.100 M AgNO 3 are mixed with 35 mL of 0.050 M K 2 CrO 4. Find the concentration of each ion in solution at equilibrium if k sp of Ag."— Presentation transcript:

1 1 Example 25 mL of 0.100 M AgNO 3 are mixed with 35 mL of 0.050 M K 2 CrO 4. Find the concentration of each ion in solution at equilibrium if k sp of Ag 2 CrO 4 = 1.1x10 -12. Solution 2 Ag 2+ + CrO 4 2-  Ag 2 CrO 4(s) mmol Ag + = 0.100 x 25 = 2.5 mmol CrO 4 2- = 0.050 x 35 = 1.75 From the stoichiometry we know that two moles of silver react with one mole of chromate, and chromate will be in excess, therefore:

2 2 I prefer that you work out the concentration of chromate left in as follows: mmol CrO 4 2- left = initial mmol CrO 4 2- - mmol CrO 4 2- reacted mmol CrO 4 2- left = initial mmol CrO 4 2- - ½ mmol Ag + mmol CrO 4 2- left = 0.05*35 – ½*0.1*25 = 0.5

3 3 mmol CrO 4 2- excess = 1.75 – 1.25 = 0.50 [CrO 4 2- ] excess = mmol/mL = 0.5/60 = 0.0083 M [NO 3 - ] = mmol/mL = 0.100 x 25/60 = 0.0417 M [K + ] = mmol/mL = 2 x 0.050 x 35/60 = 0.0583 M The silver was consumed in the reaction and the only way to calculate its concentration is through the solubility product:

4 4 K sp = [Ag + ] 2 [CrO 4 2- ] 1.1x10 -12 = (2s) 2 * (0.0083 + s) However, s is very small since the equilibrium constant is very small. Therefore, assume 0.0083>>s 1.1x10 -12 = (2s) 2 x 0.0083 s = 5.7 6 x10 -6 M Relative error = (5.7 6 x10 -6 /0.0083) x 100 = 0.07% [Ag + ] = 2s = 2x5.7 6 x10 -6 = 1.1 5 x10 -5 M [CrO 4 2- ] = 0.0083 + s = 0.0083 + 5.7 6 x10 -6 = 0.0083 M

5 5 Solubility in Presence of Diverse Ions As expected from previous information, diverse ions have a screening effect on dissociated ions which leads to extra dissociation. Solubility will show a clear increase in presence of diverse ions as the solubility product will increase. Look at the following example: Example Find the solubility of AgCl (k sp = 1.0 x 10 -10 ) in 0.1 M NaNO 3. The activity coefficients for silver and chloride are 0.75 and 0.76, respectively.

6 6 Solution AgCl (s)  Ag + + Cl - We use activities instead of concentrations since the solution contains diverse ions where we have: K sp, Th = a Ag + a Cl - K sp, Th = [Ag + ] f Ag + * [Cl - ] f Cl - 1.0x10 -10 = s x 0.75 x s x 0.76 s = 1.3x10 -5 M

7 7 The solubility of AgCl in pure water can be calculated to be 1.0x10 -5 M. If we compare this value to that obtained in presence of diverse ions we see %increase in solubility = {(1.3x10 -5 – 1.0x10 -5 )/1.0x10 -5 }x 100 = 30% Therefore, once again we have an evidence for an increase in dissociation or a shift of equilibrium to right in presence of diverse ions.

8 8 Acid-Base Equilibria

9 9 In this chapter, we study equilibrium problems as related to acids and bases as well as their salts. The main point is to know how to calculate the hydrogen ion concentration, [H + ]. However, let us start with definitions of acids and bases and then look at their equilibria.

10 10 Acid-Base Theories Four main attempts to define acids and bases are common in the literature of chemistry. Development of these attempts or theories usually followed a desire to explain the behavior of substances and account for their properties as related to having acidic or basic characteristics. Theories of acidity or basicity can be outlined below from oldest to most recent:

11 11 1. Arrhenius theory: This theory is limited to water as a solvent where an acid is defined as a substance which ionizes in water and donates a proton. A base is a substance that ionizes in water to give hydroxide ions. The hydrogen ion reacts with water to give a hydronium ion while the base reacts with water to yield a hydroxide ion. HA + H 2 O  H 3 O + + A - B + H 2 O  BH + + OH -

12 12 2. Franklin Theory: This theory introduced the solvent concept where an acid was defined as a substance that reacts with the solvent to produce the cation of the solvent. The base is a substance that reacts with the solvent to yield the anion of the solvent. HA + EtOH  EtOH 2 + + A - B + EtOH  BH + + EtO -

13 13 3. Bronsted-Lowry Theory: Some solvents like hexane or benzene are non ionizable and the Franklin theory can not be used to explain acidic or basic properties of substances. In Bronsted- Lowry theory, an acid is defined as a substance that can donate a proton while a base is a substance that can accept a proton. Also, an acid is composed of two components; a proton and a conjugate base. For example HOAc  H + + OAc - Acetic acid is an acid which donates a proton and its proton is associated with a base that can accept the proton; this base is the acetate.

14 14 4. Lewis Theory: Lewis introduced the electronic theory for acids and bases where in Lewis theory an acid is defined as a substance that accepts electrons while a base is a substance that donates electrons. Therefore, ammonia is a base because it donates electrons as in the reaction H + + :NH 3  H:NH 3 + AlCl 3 is an acid because it accepts electrons from a base such as :OR 2 AlCl 3 + : OR 2 = Cl 3 Al:OR 2

15 15 Acid-Base Equilibria in Water Fortunately, we will only deal with aqueous solutions which means that water will always be our solvent. Water itself undergoes self ionization as follows 2 H 2 O  H 3 O + + OH - K = [H 3 O + ][OH - ]/[H 2 O] 2 However, only a very small amount of water does ionize and the overall water concentration will be constant.

16 16 Therefore, one can write K w = [H 3 O + ][OH - ] K w is called autoprotolysis constant of water or ion product of water, we will also refer to [H 3 O + ] as simply [H + ] although this is not strictly correct due to the very reactive nature of H + K w = [H + ][OH - ] = 10 -14 at 25 o C.

17 17 The pH Scale In most cases, the hydrogen ion concentration is very small which makes it difficult to practically express a meaningful concept for such a small value. Currently, the pH scale is used to better have an appreciation of the value of the hydrogen ion concentration where: pH = - log [H + ] We also know that k w = [H + ][OH - ] = 10 -14 or pH + pOH = 14 Therefore, calculation of either pH or pOH can be used to find the other.


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