The Common-Ion Effect

1. The Common-Ion Effect Prepared by Prof. Odyssa Natividad R.M. Molo

2. Consider a solution that contains not only a weak acid (HC2H3O2) but also a soluble salt (NaC2H3O2) of that acid. NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq) HC2H3O2(aq)  H+(aq) + C2H3O2-(aq) • What happens when salt is added? (Remember Le Chatelier’s Principle) • Shifts equilibrium • pH of solution increases • [H+] decreases

3. The Common-Ion Effect • “The dissociation of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion common with the weak electrolyte” • Plays an important role in determining the pH of the solution & the solubility of a slightly soluble salt • Despite distinctive name, this is simply a special case of Le Chatelier’s principle

4. Sample Problem • What is the pH of a soln made by adding a 0.30 mol acetic acid & 0.30 mol sodium acetate to enough water to make 1.0L of soln? • STEPS: • Identify the major species in the soln & consider their acidity & basicity • Identify the important eqlbmrxn • Calculate the initial & eqlbmconc of each species that participate in the eqlbm • Calculate the pH from the eqlbmconc

5. Practice Exercise • Calculate the pH of a solution containing 0.085M nitrous acid (Ka = 4.5 x 10-4) & 0.10M potassium nitrite. What would be its pH if no salt were present. • Calculate the pH of a solution containing 0.20 M acetic acid & 0.30M sodium acetate. What would be the pH if no salt were present.

6. BUFFERS

7. Buffered Solution/Buffers • Solution that resist a drastic change in pH upon addition of small amounts of acid or base • Contain weak conjugate acid-base pairs • Example: • Human blood: bicarbonate-carbonic acid buffer • Normal pH: 7.35 – 7.45 • Acidosis: condition when pH falls below 7.35 • Alkalosis: when pH rises above 7.45 • Death may result if blood pH < 6.8 or > 7.8

8. Composition of Buffers • A buffer resist in pH because it contains both an acidic specie to neutralize OH- ions & a basic one to neutralize H+ ions. The acidic & basic species that make up the buffer, however, must not consume each other through a neutralization rxn. These requirements are fulfilled by a weak acid-base conjugate pair.

9. Composition cont…. • Buffers are often prepared by mixing a WA or a WB with a salt of that acid or base. • Example: • HC2H3O2 - C2H3O2- buffer (NaC2H3O2 & HC2H3O2) • NH4+ - NH3 buffer (NH4Cl & NH3) • By choosing appropriate components & adjusting their relative concentration, one can buffer a solution at virtually any pH

10. How Buffer Works • Ex: Buffer composed of weak acid (HX) & its salt (MX) HX(aq)  H+(aq) + X-(aq) • If OH- ions are added, it reacts with the acid component of the buffer to produce water & its base component (X-) OH-(aq) + HX(aq)  H2O(l) + X-(aq) • Result: [HX] dec & [X-] inc

11. How Buffer Works cont… • If H+ ions are added, it reacts with the base component of the buffer to produce water & its acid component (HX) H+(aq) + X-(aq)  HX(aq) OR H3O+(aq) + X-(aq)  H2O(l) + HX(aq) • Result: [HX] inc & [X-] dec

12. Buffer pH HX(aq)  H+(aq) + X-(aq) so • pH is determined by 2 factors: • (1) value of Ka • (2) ratio of conjugate acid-base pair • If [HX] = [X-], [H+] = Ka, pH = pKa • Result: select a buffer whose acid form has a pKa close to desired pH

13. 2 important characteristics of buffer: • (1) its capacity & (2) its pH • Buffer capacity • Is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree • Depends on the amount of Acid & Base from which the buffer is made • pH of buffer depends on • The Ka for the acid & on the relative concentration of A & B that comprise the buffer

14. To solve buffer pH • Use the same procedure to treat the common-ion effect OR the Henderson-Hasselbalch equation • Sample Problem: Calculate the pH of a buffer that is 0.12M lactic acid (Ka = 1.4 x 10-4) & 0.10M sodium lactate. • Practice: Calculate the pH of a buffer composed of 0.12M benzoic acid, HC7H5O2, (Ka = 6.3 x 10-5) & 0.20M sodium benzoate

15. Addition of SA/SB to Buffers • When a SA is added, the H+ is consumed by X- to produce HX; • Result:[HX] inc & [X-] dec • When a SB is added, the OH- is consumed by HX to produce X- • Result: [HX] dec & [X-] inc

16. Steps to calculate pH • Identify the neutralization rxn (Strong Acid & Weak Base or Strong Base & Weak Acid) • Analyze and set-up condition before & after neutralization • Calculate pH based on what is left during equilibrium condition.

17. Sample Exercise • A buffer is made by adding 0.300 mol acetic acid & 0.300 mol sodium acetate to enough water to make 1.00L soln. Calculate its pH (a) at the start; (b) after 0.020 mol KOH is added; (c) after 0.010 mol HCl is added. For comparison, what is the the pH of (d) 0.020 mol NaOH and (e) 0.010 mol HCl added in 1.00L pure water.

18. Practice Exercise • What are the effects on the pH of adding (0.0060 mol HCl and (b) 0.0060 mol NaOH to 0.300L of a buffer solution that is 0.250M acetic acid (Ka = 1.8 x 10-5) & 0.560 M sodium acetate. For comparison, what is the pH of (c) 0.0060 mol HCl & (d) 0.0060 mol NaOH in 0.300L water. • A 1.00 L volume of buffer is made with concentrations of 0.350 M NaCHO2 (sodium formate) and 0.550M HCHO2 (Ka = 1.8 x 10-4). (a) What is the initial pH? (b) What is the pH after the addition of 0.0050 mol HCl? (assume that the volume remains 1.00L) (c) What would be the pH after the addition of 0.0050 mol NaOH to the original buffer?