CHEM 1011 pH and Buffer Solutions

Brønsted-Lowry Theory Acid-proton donor Base-proton acceptor

pH and Buffer Solutions For example, HCl + H 2 O ↔ H 3 O + + Cl - For the forward reaction, HCl – proton donor - acid H 2 O – proton acceptor - base

pH and Buffer Solutions For example, HCl + H 2 O ↔ H 3 O + + Cl - For the reverse reaction, H 3 O + – proton donor - acid Cl - – proton acceptor - base

pH and Buffer Solutions According to Brønsted-Lowry Theory, every acid-base reaction creates conjugate acid-base pair.

pH and Buffer Solutions For example, HCl + H 2 O ↔ H 3 O + + Cl - HCl - acid, Cl - - conjugate base H 2 O - base, H 3 O + - conjugate acid

pH and Buffer Solutions Acids can be monoprotic, diprotic, or triprotic.

pH and Buffer Solutions Monoprotic – donate one proton e.g., HCl, HNO 3, HCOOH, CH 3 COOH HCl + H 2 O ↔ H 3 O + + Cl - HNO 3 + H 2 O ↔ H 3 O + + NO 3 - HCOOH + H 2 O ↔ H 3 O + + HCOO - CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -

pH and Buffer Solutions Diprotic – donate two protons e.g, H 2 SO 4, H 2 CO 3 H 2 SO 4 + H 2 O ↔ H 3 O + + HSO 4 - HSO H 2 O ↔ H 3 O + + SO 4 2- H 2 CO 3 + H 2 O ↔ H 3 O + + HCO 3 - HCO H 2 O ↔ H 3 O + + CO 3 2-

pH and Buffer Solutions Triprotic – donate three protons e.g, H 3 PO 4 H 3 PO 4 + H 2 O ↔ H 3 O + + H 2 PO 4 - H 2 PO H 2 O ↔ H 3 O + + HPO 4 2- HPO H 2 O ↔ H 3 O + + PO 4 3-

pH and Buffer Solutions Water can self ionize. H 2 O + H 2 O ↔ H 3 O + + OH - [H 3 O + ] = 1 x moles/L [OH - ] = 1 x moles/L K w = [H 3 O + ] [OH - ] = [1 x ] [1 x ] = 1 x

pH and Buffer Solutions K w same for all aqueous solutions. Can calculate [OH - ] for solution of strong acid. e.g., 0.05 M HNO 3 [H 3 O + ] = 5 x moles/L [OH - ] = K w / [H 3 O + ] [OH - ] = 1 x / 5 x = 2 x mol/L

pH and Buffer Solutions Can also calculate [H 3 O + ] of solution of strong base. e.g., 0.04 M NaOH [OH - ] = 4 x mol/L [H 3 O + ] = K w / [OH - ] [H 3 O + ] = 1 x / 4 x = 2.5 x

pH and Buffer Solutions pH scale goes from 0 to 14 pH < 7 acidic pH = 7 neutral pH > 7 basic

pH and Buffer Solutions pH = - log [H 3 O + ] e.g., pH of 0.05 M HNO 3 (a strong acid) pH = - log [5 x ] = 1.12

pH and Buffer Solutions The pOH of the solution of a strong base can also be calculated and used to determine pH. e.g., 0.04 M NaOH pOH = -log[OH - ] = -log[4 x ] = 1.10 pH + pOH = pK w = 14 pH = 14 - pOH = = 12.9

pH and Buffer Solutions You will use pH paper to measure the pHs of your solutions. Although a pH meter would yield more accurate results, it is unnecessary.

pH and Buffer Solutions Buffer solutions Resist changes of pH following the addition of acid or base. Combinations of weak acids and their conjugate bases.

pH and Buffer Solutions Buffered aspirin Blood buffers prevent acidosis and alkalosis. H 2 CO 3 + H 2 O ↔ HCO H 3 O + H 2 PO H 2 O ↔ HPO H 3 O +

pH and Buffer Solutions The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer system. pH = pK a + log [A - ] / [HA] pH = pK a when [A - ] = [HA] Buffer activity continues as long as log [A - ] / [HA] ratio kept constant

Conclusions and Calculations Answer Post-Lab questions 1 and 3 based upon your observations (omit 2).

Conclusions and Calculations For Post-Lab question 4, the HCl reacts with the CH 3 COO -, and the NaOH reacts with the H 3 O + An acetic acid-acetate buffer system would be represented by CH 3 COOH + H 2 O ↔ CH 3 COO - + H 3 O +

Conclusions and Calculations For an acetic acid-acetate buffer system pH = pK a + log [CH 3 COO - ] / [CH 3 COOH] pK a = [CH 3 COO - ][H 3 O + ] / [CH 3 COOH] = 4.75 pH = log [CH 3 COO - ] / [CH 3 COOH] Use this relationship when answering Pre-Lab question 5 and Post-Lab questions 5 a. and b.

Conclusions and Calculations For Post-Lab questions 5 c. and d., H 2 CO 3 + H 2 O ↔ HCO H 3 O + pH = pK a + log [HCO 3 - ] / [H 2 CO 3 ] pK a = [HCO 3 - ][H 3 O + ] / [H 2 CO 3 ] = 6.37 pH = log [HCO 3 - ] / [H 2 CO 3 ]