School of Chemistry, Howard College Campus University of KwaZulu-Natal CHEMICAL ENGINEERING CHEMISTRY 2 (CHEM171)

School of Chemistry, Howard College Campus University of KwaZulu-Natal CHEMICAL ENGINEERING CHEMISTRY 2 (CHEM171) Lecture Notes 1 st Series: Solution Chemistry of Salts SALTS Preparation Note, an acid is a proton donor and a base is a proton acceptor. A salt arises from the reaction of an acid with a base. For instance the reaction between HBr and Mg(OH) 2 leads to MgBr 2. 2HBr(aq) + Mg(OH) 2 (aq) MgBr 2 (aq) + 2H 2 O(l) Question 1.1 Write balanced chemical equations for the formation of the following salts: (a) sodium fluoride (b) magnesium carbonate (c) aluminium sulfate (d) calcium phosphate Dissolution When a solid sample of a salt (the solute) is placed in water (the solvent) the salt dissolves to give a solution. For instance NaCl(s) + (aq) Na + (aq) + Cl - (aq) solute solvent solution We now have a solution which contains ions. If we place two electrodes into this solution and apply a potential difference between these electrodes (i.e. make one electrode positive and the other negative) then the cations in solution will migrate towards the negative electrode and the anions will migrate towards the positive electrode. power source +ve -ve This movement of charge within the solution causes a current to flow and the substance which supplied the ions (in this case sodium chloride) is said to be an electrolyte. Electrolytes can be further divided into strong electrolytes and weak electrolytes. A strong electrolyte is one that fully dissociates into ions in solution. All salts (except for a few cadmium, mercury and lead salts, which we won't be meeting in this course) are strong electrolytes. 1

SALTS dissociate fully in water to form ionic solutions and are thus STRONG ELECTROLYTES. A weak electrolyte is a substance which does not dissociate fully into ions when in solution. When acetic acid is dissolved in water some of the molecules do not dissociate to form ions. Rather an equilibrium is established. CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) A measure of how many acetic acid molecules dissociate can be obtained from the K a value. Example 1.1 What percentage of molecules have dissociated in a 0.100 mol dm -3 solution of acetic acid given K a (CH 3 COOH) = 1.75 x 10-5? All weak acids and weak bases are weak electrolytes. Question 1.2 Classify each of the following species as either a strong electrolyte or a weak electrolyte. (a) HF (b) HCl (c) HNO 3 (d) CH 3 CH 2 COOH (e) NH 3 (f) NaOH (g) H 2 O Question 1.3 Which would give you the larger current in a conductivity cell, 0.10 mol dm -3 solution of CH 3 COOH (K a = 1.75 x 10-5 ) or 0.10 mol dm -3 solution of HCN (K a = 6.2 x 10-10 )? Explain your answer clearly and quantitatively. 2

Precipitation Reactions What happens when we add a solution of a salt to a solution of another salt? The answer is IT DEPENDS! If for instance we add the two following salts to each other some students might write KNO 3 (aq) + NaCl(aq)? KNO 3 (aq) + NaCl(aq) KCl(aq) + NaNO 3 (aq) where a replacement reaction has occurred and the cations have swopped over anions. There is of course a rather large problem with this equation. If we write it out more fully Initial: Final: K + (aq) + NO - 3 (aq) + Na + (aq) + Cl - (aq) K + (aq) + Cl - (aq) + Na + (aq) + NO - 3 (aq) Therefore nothing has happened! There is no reaction. A reaction will only happen between salts in solution if ions are removed from solution. There are two important ways ions can be removed from solution either through formation of a weak electrolyte or, far more commonly, through formation of a precipitate. How do we know which salts are likely to precipitate out of solution? Well, the simple answer is those which are not soluble in water. However, predicting the solubility of a salt tends to be rather an inexact science in chemistry. There are certain rules of thumb that you should have in mind when determining what will happen during a reaction between salts. Salts containing Group 1 metal ions are soluble Salts containing the NH 4 + ion are soluble Nitrates and acetates are soluble Chlorides, bromides and iodides are soluble except those of Ag +, Pb 2+ and Hg 2 2+ Sulfates are soluble except those of Pb 2+, Hg 2 2+, Sr 2+ and Ba 2+. Ag 2 SO 4 and CaSO 4 are only slightly soluble Carbonates, phosphates and sulfites are insoluble (except those of group 1 andnh 4 + ). Sulfides (S 2- ) are insoluble except those of Groups 1 and 2 and NH 4 + If, therefore, we do the following reaction BaCl 2 (aq) + Na 2 SO 4 (aq)? then, this time a reaction will take place as when the Ba 2+ ions meet the SO 4 2- ions in solution they will form a precipitate and be removed from solution BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl(aq) leaving behind a solution of sodium chloride. Question 1.4 Predict the product (if any) from the following reactions by writing balanced chemical equations. (a) AgNO 3 (aq) + NaCl(aq) (b) sodium phosphate and magnesium chloride (c) lead(ii) chloride and potassium bromide (d) ammonium chloride and sodium carbonate. 3

Sparingly Soluble Salts There is a more quantitative way to measure the solubility of salts. In the previous section we either considered a salt to be soluble or non-soluble. The truth of the matter is that these are the two extremes of a continuum. A more exact measure could be given if we look at the equilibrium between a precipitate and its constituent ions in solution. If we take BaSO 4 as an example: BaSO 4 (s) + (aq) Ba 2+ (aq) + SO 4 2- (aq) The equilibrium constant for this type of equation is known as the solubility product, K sp. K sp = [Ba 2+ ][SO 4 2- ] Note: We ignore concentration terms for pure solids and liquids (as well as the term for the solvent) in solution equilibrium constant expressions. For BaSO 4 K sp = 1.1 x 10-10. If the value of K sp is very small (as for BaSO 4 ) then that means the ionic product ([Ba 2+ ] x [SO 4 2- ] in this case) is very small which in turn means there are very few ions in solution. Salts which have a K sp value (i.e. those which are not fully soluble in water) are said to be sparingly-soluble. Calculating Molar Solubilities K sp values are important in that we can calculate the molar solubility of a salt in water from the K sp value. Firstly though we should make sure everyone understands the meaning of K sp values. If we take 1 gram of AgCl and place it in 100 g of water about 1 x 10-4 g of the salt dissolves. If we then take 10 gram of AgCl and place it in 100 g of water, again, only 1 x 10-4 g of the salt dissolves. At equilibrium the solution is said to be saturated. The product of the concentrations of ions in solution cannot exceed the ionic product constant. Thus the solubility of a salt is independent of how much of the solid is present. As long as some solid is present an equilibrium is established and the K sp value is a constant at a given temperature. Example 1.2 Calculate the molar solubility of AgCl at 25 o C, given the relevant K sp value is 1.82 10-10. When the stoichiometry of the salt is not 1:1 then we have to be a little more careful. 4

Example 1.3 Calculate the molar solubility of PbCl 2 given the relevant K sp value is 1.7 x 10-5. Question 1.5 Calculate the molar solubilities of the following salts (a) TlCl (K sp = 1.8 x 10-4 ) (b) Ag 3 AsO 4 (K sp = 6 x 10-23 ) (c) Ag 2 S (K sp = 8 x 10-51 ) Question 1.6 A saturated solution of PbI 2 at 25 o C has a Pb 2+ molar concentration of 1.3 x 10-3 mol dm -3. Calculate K sp for PbI 2. The Common Ion Effect In reality the calculations in the previous section have very little meaning when we look at solubilities of solids in the environment. The trouble with the environment is that it is really quite a complicated place. There are many different solids present and the equilibria that each establishes has an effect on all the other equilibria. Consider for example if we have both NaCl and AgCl dissolved in the same body of water. Let s imagine that the dissolution of NaCl, being freely soluble, results in a chloride ion concentration of 0.100 mol dm -3. How will this affect the solubility of AgCl? K sp = [Ag + ] [Cl - ] = 1.82 x 10-10 This time however the two ions are not present at the same concentration. The chloride ion concentration has two inputs the 0.100 mol dm -3 arising from the NaCl plus the Cl - ions arising from AgCl. 5

If we set the molar solubility of [AgCl] as S mol dm -3 then [AgCl] = [Ag + ] = S mol dm -3 and [Cl - ] = (0.100 + S ) mol dm -3 Now we can substitute these concentrations into the K sp expression. K sp = [Ag + ] [Cl - ] = S (0.100 + S ) S 2 + 0.100 S = 1.82 x 10-10 and we need to solve the quadratic S = -b ± b 2 2a -4ac -0.100 ± 0.0100+ 4 1.82 10 = 2 S = 1.82 10-9 [Ag + ] = [AgCl] = 1.82 x 10-9 mol dm -3-10 We could of course save ourselves from having to solve a quadratic by using a bit of the old CCS. CCS? you ask Chemical Common Sense If we consider the two sources of chloride in the solution it is obvious that one is much bigger than the other. When considering the Cl - concentration as 0.100 + S we should realise that in all likelihood 0.100 >> S Therefore (0.100 + S ) simplifies to 0.100 and K [AgCl] = [Ag + sp ] = [Cl ] 10 1.82 10 = 0.100 = 1.82 x 10-9 mol dm -3 This is exactly what we got from solving the quadratic. Example 1.4 Calculate the molar solubility of PbI 2 (K sp = 7.9 10-9 ) in a solution containing 0.214 mol dm -3 iodide ions. 6

Question 1.7 Calculate the molar solubility of lanthanum iodate, La(IO 3 ) 3, (K sp = 1.0 10-11 ) in a solution which contains 0.124 mol dm -3 of potassium iodate(v). If we now look more closely at two of the numbers that we ve just calculated, Molar solubility mol dm -3 AgCl in H 2 O 1.35 x 10-5 AgCl in 0.100 M Cl - 1.82 x 10-9 then it becomes apparent that the molar solubility of a sparingly-soluble salt decreases when the solution contains a common ion. A common ion is one which is present both in the solution before the addition of the solid and in the solid itself. Therefore the chloride ion is a common ion in the AgCl/NaCl example. Of course we could have predicted this before we did any calculations! If we look again at the equilibrium involved when AgCl dissolves AgCl(s) Ag + (aq) + Cl - (aq) then we can use Le Chatelier s principle to explain why AgCl becomes less soluble in the presence of an external source of Cl - ions. If we add Cl - to this equilibrium then the position of the equilibrium shifts to the left and the solubility of AgCl (as measured by the Ag + concentration) decreases. The Common Ion Effect results in the lowering of the molar solubilities of sparingly-soluble salts in solution. Hydrolysis of Salts Although salts might react with each other, as we saw in the previous section, they might also react with water itself. This is called HYDROLYSIS. Hydrolysis is very important because it invariably results in the ph of the solution changing. Two types of hydrolysis can occur. (1) MX(s) + (aq) M + (aq) + X (aq) X (aq) + H 2 O(l) XH(aq) + OH (aq) Here the anion is acting like a base. It accepts H + from water. Examples would be the complex anions SO 4 2-, CO 3 2-, and C 2 O 4 2- (the oxalate ion). Question 1.8 Will the ph decrease or increase during this type of hydrolysis? The other type of hydrolysis can occur if we have a complex cation which can act as an acid and donate a proton to water (2) HMX(s) + (aq) HM + (aq) + X (aq) HM + (aq) + H 2 O(l) M(aq) + H 3 O + (aq) An example of the above would be NH 4 +, the ammonium ion. Question 1.9 Write down the two relevant equations when we dissolve ammonium chloride in water. Is this an acidic or basic solution? 7

The ph of salt solutions We know that a salt arises from the reaction of an acid with a base. Some typical salts are sodium ethanoate (ethanoate = acetate), magnesium bromide, potassium chloride and calcium carbonate. There is a very easy way to determine whether a salt gives rise to a neutral, basic or acidic solution when dissolved in water, and that is to recognise which acid and base were used to prepare the salt. (i) Salts prepared from a strong acid and a strong base If we mix a stoichiometric amount of sodium hydroxide and hydrochloric acid together, for instance, then a solution containing sodium chloride will result. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Note: this is not an equilibrium, effectively, the reaction goes to completion. Neither Na + (aq) nor Cl - (aq) undergo any reaction with water. Therefore there is no excess H + or excess OH - generated in solution. The ph is therefore 7.00 at 25 o C. This is true for all salts prepared from the addition of strong acids to strong bases. Salts prepared from reacting strong acids with strong bases all have a ph equal to 7.00 at 25 o C. (ii) Salts prepared from strong acids and weak bases If we take HCl, a strong acid, and react it with NH 3, a weak base, in the stoichiometric ratio, then ammonium chloride will result. Is this salt acidic, neutral or basic? If we dissolve NH 4 Cl(s) in water then a solution containing ammonium ions and chloride ions results. NH 4 Cl(s) + (aq) NH + 4 (aq) + Cl - (aq) This is not an equilibrium; all the NH 4 Cl will dissolve as all ammonium salts are soluble. However NH + 4 (aq) is an acid. The following equilibrium will be established in solution. NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) If hydronium ions are generated in this hydrolysis reaction then the solution is obviously acidic, i.e. ammonium ions are acidic as they generate H 3 O + ions in solution. To calculate the ph of an acidic salt (i.e. a weak acid) we need to know two things the K a and C a (or more exactly C s, the concentration of the salt). We have seen previously that the approximation holds for cases where C a > 100 K a. [H + ] = K a C a Example 1.5 Calculate the ph of a 0.0924 mol dm -3 solution of NH 4 Cl at 25 o C. 8

Salts prepared from reacting a strong acid with a weak base all have a ph < 7.00 as the cation of the salt undergoes hydrolysis to form H 3 O +. Question 1.10 Calculate the ph of a solution of 0.1046 mol dm -3 of NH 4 Br(aq). Question 1.11 Which is more acidic; a 0.100 mol dm -3 NH 4 Cl solution or a 0.100 mol dm -3 solution of NH 4 Br? (iii) Salts prepared from a weak acid and a strong base If we take a strong base, such as NaOH, and react it with the stoichiometric amount of a weak acid such as CH 3 COOH, then we end up with sodium ethanoate (or sodium acetate as it is more commonly known). Is sodium acetate acidic, neutral or basic? If we take a solid sample of sodium acetate and dissolve it in water then NaOOCCH 3 (s) + (aq) Na + (aq) + CH 3 COO - (aq) This time it is the anion we should keep our eyes on. The acetate ion will react with water thus CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) As hydroxide ions are generated in solution, the acetate ion is basic. For a weak base we can calculate the ph quite simply using the equation. [OH - ] = K b C b so long as C b > 100 K b Question 1.12 If K a for CH 3 COOH is 1.75 x 10-5, what is K b for CH 3 COO -? Example 1.6 Calculate the ph of a 0.154 mol dm -3 solution of sodium acetate. Salts prepared from reacting a weak acid with a strong base all have a ph > 7.00, as the anion of the salt undergoes hydrolysis. Question 1.13 Classify each of the following salts as acidic, neutral or basic. (a) potassium iodide 9

(b) potassium propanoate (K + - OOCCH 2 CH 3 ) (c) potassium perchlorate (d) potassium chlorate (V) (e) ammonium nitrate (iv) Salts prepared from a weak acid and a weak base If we take a salt such as ammonium fluoride, then predicting whether the solution is acidic or basic is not as straightforward as for the preceding cases. Here we have prepared the salt from HF(aq) + NH 3 (aq) NH 4 F(aq) weak acid weak base Both the cation and the anion of the salt can now undergo hydrolysis. NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) F - (aq) + H 2 O(l) HF(aq) + OH - (aq) The overall ph will depend upon the relative strength of the weak acid (NH 4 + ) and the weak base (F - ). This can be gauged from the relevant K a and K b values. These calculations are not within the scope of this course, however. Instead simply realise that salts prepared from mixing weak acids and weak bases can be acidic, neutral or basic! Buffers Buffers are what keep us alive. Our bodies (our internal environments if you like) are dominated by the buffer action of our blood. Our external environments, the waters and soils upon which we live, are also dominated by buffers. So what is a buffer? A buffer is a solution which resists changes in its ph. To resist changes in ph buffers require two components; something which will react with a strong base and something else which will react with a strong acid. If we have a system which can mop up H 3 O + and OH - ions then we have a buffer. An alternative definition of a buffer is a solution which can maintain its ph (approximately) upon addition of a strong acid or a base. Buffers come in two forms - acidic buffers, solutions which maintain a ph < 7 and basic buffers, solutions which can maintain a ph > 7. If you re lucky you can even have a buffer which maintains a neutral ph! Buffers of ph < 7.00 Buffers which maintain an acidic ph contain two components a weak acid and the salt of a weak acid. An example would be: WEAK ACID + SALT OF A WEAK ACID CH 3 COOH + CH 3 COO - Na + Acetic acid sodium acetate The weak acid is there to react with any OH - ions which might be introduced into the system. CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) The effective result of this reaction is to convert a strong base (OH - ) into a weak base (CH 3 COO - ). Thus although the ph will increase due to more base being present, the increase will be relatively small as the acetate ion is a weak base (K b = 5.7 10-10 ). The salt of the weak acid is there to react with any H 3 O + ions which might be introduced into the system. 10

CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O(l) Again it is the effective result of this reaction which is important- a strong acid (H 3 O + ) has been converted into a weak acid (K a (CH 3 COOH) = 1.75 10-5 ). The ph of the solution will decrease but only by a small amount. Question 1.14 Calculate the equilibrium constant for the following reactions at 25 o C. (a) (b) CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O(l) This buffer solution can be described by the following equilibrium CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) [CH COO ][H O ] 3 3 and K a = [CH COOH] If we now take ( log) of both sides of the equation 3 - + pk a = ph log [CH COO ] 3 [CH COOH] 3 - as - log K a = pk a - log [H + ] = ph and when we make ph the subject of the equation, we get a most important equation [CH3COO ] ph = pk a + log [CH COOH] 3 - This type of equation is so important because it allows us to calculate the ph of any buffer solution. If we define CH 3 COO - as a proton acceptor i.e. a base, and CH 3 COOH as a proton donor i.e. an acid, then the Henderson-Hasselbalch (the H-H) equation appears. ph = pk a + log [proton acceptor] [proton donor] Henderson- Hasselbalch equation Example 1.7 Calculate the ph of a solution which contains 0.1046 mol dm -3 of CH 3 COOH and 0.1247 mol dm -3 of sodium acetate. Normally, acidic buffers are prepared by adding a strong base to a weak acid until the desired ratio of proton acceptor/proton donor is reached. 11

Example 1.8 Calculate the ph of the solution prepared from adding 50.00 ml of 0.1011 mol dm -3 NaOH to 100.0 ml of a 0.1122 mol dm -3 CH 3 CH 2 COOH solution (pk a CH 3 CH 2 COOH = 4.88). It is important that you always take time to realize what is present in solution. Once it is ascertained that a buffer solution is present we can use the Henderson-Hasselbalch Equation Question 1.15 Calculate the ph of the solution resulting from the addition of 25.00 ml of 0.0924 mol dm -3 NaOH to 50.00 ml of 0.1045 mol dm -3 CH 3 COOH (pk a CH 3 COOH = 4.76). Question 1.16 Calculate the ph of the solution resulting from the addition of 75.00 ml of 0.0924 mol dm -3 NaOH to 50.00 ml of 0.1045 mol dm -3 CH 3 COOH. 12

Buffers of ph > 7.00 Basic buffers contain TWO species a weak base and the salt of a weak base. For example WEAK BASE + SALT OF A WEAK BASE NH 3 + NH 4 Cl ammonia + ammonium chloride The weak base will take care of any added H 3 O +. NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O(l) Again we are effectively turning a strong acid into a weak acid (NH 4 + ). The salt of a weak base will react with any added OH - NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O(l) and the strong base gives way to a weak base (NH 3 ). Question 1.17 If K a (NH 4 + ) = 5.71 10-10, calculate the equilibrium constant for each of the two reactions below. (a) NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O(l) (b) NH + 4 (aq) + OH - (aq) NH 3 (aq) + H 2 O(l) The equilibria which determines the final ph of the buffer will be NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) The H-H equation will therefore look like this. ph = pk a + log [NH [NH 3 + 4 ] ] If instead we had written the equilibrium as NH 3 (aq) + H 2 O(l) NH + 4 (aq) + OH - (aq) and therefore [NH4][OH ] then K b = [NH ] + 3 - poh = pk b + log [NH [NH + 4 3 ] ] Both forms of the H-H equation are equivalent after all we can t have a solution with two different values of ph. It is a matter of personal preference which form of the equation you use, just don t get them mixed up! 13

Example 1.9 Calculate the ph of a solution prepared from 100.0 ml of 0.2041 mol dm -3 NH 3 and 100.0 ml of 0.3267 mol dm -3 NH 4 Cl (K a (NH 4 + ) = 5.71 10-10 ). Question 1.18 Derive the H-H equation for a buffer solution prepared from mixing HCN and NaCN. Question 1.19 Calculate the ph of a solution prepared by mixing 250 ml of 0.1113 mol dm -3 HCl with 450 ml of 0.0925 mol dm -3 NH 3. The effect of ph on molar solubilities Chemistry has a habit of coming full circle. We have dealt with the precipitation and then the hydrolysis of salts. Both reaction types involve equilibria. Are these equilibria linked to each other? If we think about it carefully the answer has to be YES. Consider again the salt barium sulfate BaSO 4 (s) + (aq) Ba 2+ (aq) + 2 SO 4 (aq) Can either of these ions in solution hydrolyse? If so then the position of the solubility equilibria above will be affected. This can sometimes be a difficult topic to grasp so we ll start with a special group of compounds, the sparinglysoluble hydroxides, to illustrate this point clearly and then we ll return to the salts. 14

Magnesium hydroxide has a K sp of 7.1 x 10-12. Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) If we decrease the ph of the solution, say by adding an acidic buffer, then what will happen to the solubility of Mg(OH 2 )? Let s break the process down into successive statements. By adding an acid we increase [H + ] and consequently decrease [OH - ] as K w must be maintained. If we decrease the amount of OH - (aq) in the system the position of equilibrium will shift to the right. If the position of equilibrium moves to the right then the concentration of Mg 2+ increases. As [Mg(OH) 2 ] = [Mg 2+ ], the molar solubility of magnesium hydroxide increases as the ph is lowered. The molar solubility of sparingly-soluble hydroxides increases as the ph is decreased. Example 1.10 Calculate the solubility of Zn(OH) 2 (K sp = 3.0 10-16 ) (a) in pure water (b) in a buffer at ph = 7.00 (c) in a buffer at ph = 11.00 Question 1.20 (a) Calculate the equilibrium ph when Zn(OH) 2 is added to water. (b) Explain the three concentrations calculated in example 2.10 by means of Le Chatelier s Principle. 15

Some sparingly-soluble salts, which are not hydroxides, also exhibit ph dependency for their molar solubilities. These are solids which give rise to ions which hydrolyse in solution. We ve met this type of ion before when dealing with weak acids and bases. Let s re-examine the solubility of Ag 2 S (see Question 1.5(c)) Ag 2 S(s) 2 Ag + (aq) + S 2- (aq) We should immediately recognise that S 2- is a base, i.e. it can accept protons from water. S 2- (aq) + H 2 O(l) HS - (aq) + OH - (aq) What therefore happens if we add a buffer to this solution which decreases the ph? Again let s work through this process stepwise. If we add an acid, [H + ] [OH - ]. If [OH - ] then the position of the equilibrium S 2- (aq) + H 2 O(l) HS - (aq) + OH - (aq) moves to the right. This results in removing S 2- (aq) from solution as HS - (aq). If S 2- (aq) is removed from solution then the position of the equilibrium Ag 2 S(aq) 2 Ag + (aq) + S 2- (aq) also moves to the right, increasing [Ag + ]. As [Ag 2 S] = ½[Ag + ] then as the ph is decreased the molar solubility of [Ag 2 S] increases. Thus, if either ion resulting from the dissolution of a solid hydrolyses with the solvent, then the molar solubility of the sparingly-soluble salt increases. Example 1.11 Would you expect the molar solubility of BaSO 4 in water to be ph dependent? Explain your answer. Question 1.21 Would you expect the molar solubility of the following classes of sparingly-soluble salts to be ph dependent? Explain your answer in each case. (a) carbonates, e.g. CaCO 3 (b) oxalates, e.g. CdC 2 O 4 (c) arsenates, e.g. Ag 3 AsO 4 16