We are faced with different types of solutions that we should know how to calculate the pH or pOH for. These include calculation of pH for 1. Strong acids and strong bases 2. Weak acids (monoprotic) and weak bases (monobasic) 3. Salts of weak acids and salts of weak bases 4. Mixtures of weak acids and their salts (buffer solutions) 5. Polyprotic acids and their salts and polybasic bases and their salts We shall also look at pH calculations for mixtures of acids and bases as well as pH calculations for very dilute solutions of the abovementioned systems.

pH calculations 1. Strong Acids and Strong Bases Strong acids and strong bases are those substances which are completely dissociated in water and dissociation is represented by one arrow pointing to right. Examples of strong acids include HCl, HNO3, HClO4, and H2SO4 (only first proton). Examples of strong bases include NaOH, KOH, Ca(OH)2, as well as other metal hydroxides.

Example Find the pH of a 0.1 M HCl solution. Solution HCl is a strong acid that completely dissociates in water, therefore we have HCl  H+ + Cl- H2O D H+ + OH- [H+]Solution = [H+]from HCl + [H+]from water However, [H+]from water = 10-7 in absence of a common ion, therefore it will be much less in presence of HCl and can thus be neglected as compared to 0.1 ( 0.1>>[H+]from water) [H+]solution = [H+]HCl = 0.1 pH = -log 0.1 = 1

We can always look at the equilibria present in water to solve such questions, we have only water that we can write an equilibrium constant for and we can write:

Kw = (0.1 + x)(x) However, x is very small as compared to 0.1 (0.1>>x) 10-14 = 0.1 x x = 10-13 M Therefore, the [OH-] = 10-13 M = [H+]from water Relative error = (10-13/0.1) x 100 = 10-10% [H+] = 0.1 + x = 0.1 + 10-13 ~ 0.1 M pH = 1

Example Find the pH of a 1x10-5 M HNO3 solution. Solution Nitric acid is a strong acid which means that it dissociates completely in water. Therefore, [H+]from Nitric acid = 1x10-5 M We can now set the equilibria in water as above

Therefore, the[OH-] = 10-9 M = [H+]from water Kw = (10-5 + x)(x) However, x is very small as compared to 10-5 (10-5>>x) 10-14 = 10-5 x x = 10-9 M Therefore, the[OH-] = 10-9 M = [H+]from water Relative error = (10-9/10-5) x 100 = 0.01 % [H+] = 10-5 + 10-9 ~ 10-5 M pH = - log 10-5 = 5

Find the pH of a 10-7 M HCl solution. Solution Example Find the pH of a 10-7 M HCl solution. Solution HCl is a strong acid, therefore the [H+]from HCl =10-7 M

Kw = (10-7 + x)(x) Let us assume that x is very small as compared to 10-7 (10-7>>x) 10-14 = 10-7 x x = 10-7 M Therefore, the[OH-] = 10-7 M = [H+]from water Relative error = (10-7/10-7) x 100 = 100 % Therefore, the assumption is invalid and we have to solve the quadratic equation. Solving the quadratic equation gives: [H+] = 7.62x10-7 M pH = 6.79

Calculate the pH of the solution resulting from mixing 50 mL of 0 Calculate the pH of the solution resulting from mixing 50 mL of 0.1 M HCl with 50 mL of 0.2 M NaOH. Solution When HCl is mixed with NaOH neutralization takes place where they react in a 1:1 mole ratio. Therefore, find mmol mixed of each reagent to see if there is an excess of either reagent mmol HCl = 0.1 x 50 = 5 mmol mmol NaOH = 0.2 x 50 = 10 mmol mmol NaOH excess = 10 – 5 = 5 mmol [OH-] = 5/100 = 0.05 M The hydroxide ion concentration is high enough to neglect the contribution from water. pOH = 1.3 and pH = 14 – pOH = 14 – 1.3 = 12.7

Find the pH of a solution prepared by mixing 2 Find the pH of a solution prepared by mixing 2.0 mL of a strong acid at pH 3.0 and 3.0 mL of a strong base at pH 10. Solution First, find the concentration of the acid and the base pH = 3.0 means [H+] = 10-3.0 pH = 10 means [H+] = 10-10 M or [OH-] = 10-4 M Now find the number of mmol of each mmol acid = 10-3 x 2.0 = 2.0x10-3 mmol mmol base = 10-4 x 3.0 = 3.0 x 10-4 mmol mmol acid excess = 2.0x10-3 – 3.0x10-4 = 1.7x10-3 [H+] = 1.7x10-3/5 = 3.4x10-4 M pH = - log 3.4x10-4 = 3.5

2. Salts of Strong Acids and Bases The pH of the solution of salts of strong acids or bases will remain constant (pH = 7) where the following arguments apply: NaCl dissolves in water to give Cl- and Na+. For the pH to change, either or both ions should react with water. However, is it possible for these ions to react with water? Let us see:

Cl- + H2O D HCl + OH- (wrong equation) Now, the question is whether it is possible for HCl to form as a product in water!! Of course this will not happen as HCl is a strong acid which is 100% dissociated in water. Therefore, Cl- will not react with water but will stay in solution as a spectator ion. The same applies for any metal ion like K+ where if we assume that it reacts with water we will get:

K+ + H2O D KOH + H+ (wrong equation) Now, the question is whether it is possible for KOH to form as a product in water!! Of course this will not happen as KOH is a strong base which is 100% dissociated in water. Therefore, K+ will not react with water but will stay in solution as a spectator ion. It is clear now that neither K+ nor Cl- react with water and the hydrogen ion concentration of solutions of salts of strong acids and bases comes from water dissociation only and will be 10-7 M (pH =7).

Weak Acids and Bases A weak acid or base is an acid or base that partially (less than 100%) dissociated in water. The equilibrium constant is usually small and, in most cases, one can use the concepts mentioned in the equilibrium calculations section discussed previously with application of the assumption that the amount dissociated is negligible as compared to original concentration. This assumption is valid if the equilibrium constant is very small and the concentration of the acid or base is high enough.