Lesson # 7 Equilibrium & pH Chemical Equilibrium Lesson # 7 Equilibrium & pH

Water as an Acid and Base We have already learned that water is amphiprotic – it can act as both an acid and a base. It can even act as an acid and base in the same reaction – the autoionization of water H2O (l) + H2O (l) ⇌ Kw = Experiments show that at 25ºC, [H+] = [OH-] = 1.0 x 10-7 mol/L

Weak Acids & Bases This relationship also applies to weak acids and bases. HA (aq) + H2O (l) ⇌ A- (aq) + H3O+ (aq) Ka = [H+][A-] [HA] A- (aq) + H2O (l) ⇌ HA (aq) + OH- (aq) for the conjugate base Kb = [HA][OH] [A-] If we add these two reactions together, we simply get: which is the same as the autoionization as water.

Weak Acids & Bases If we multiply Ka x Kb we get: Overall, this means that Ka x Kb = Kw for any weak acid and its conjugate base, and any base with its conjugate acid. This explains what we learned yesterday – as the strength of the acid increases, the strength of its conjugate base decreases.

Example 1 The hydrogen phosphate ion, HPO42- (aq), has a Ka value of 1.3x10-13 at SATP. What is the base ionization constant, Kb, for the phosphate ion, PO43- (aq)?

pH and pOH Last year you learned that pH is the negative logarithm of the hydrogen ion concentration. A logarithm is commonly used to convert very small concentration values into more convenient positive integer values. pH = The negative logarithm of the hydroxide concentration is known as the pOH. pOH = In the first section, we noted that the concentration of both [H+] and [OH-] at SATP was 1.0x10-7. pH = pOH = This is true of all neutral aqueous solutions.

pH and pOH When we acid or base to pure water, the pH or pOH changes. Adding acid, for example, would increase the concentration, which would, in term, decrease the pH (making it more acidic). It would also increase the pOH (if the hydrogen ion concentration goes up, the hydroxide goes down).

Example 2 Hydrochloric acid solution is added to water. The concentration of acid is determined to be 1.0x10-3 mol/L. Determine the pH and pOH of the solution. pH = *The number of decimal places in the log value must equal the number of significant digits in the original value.

Example 2 (continued) pOH = Therefore, pH + pOH = at SATP.

Converting from pH or pOH to Concentration [H=] = 10-pH [OH-] = 10-pOH

Example 3 A solution of ammonia has a pOH of 11.09. What is the [OH-] and [H+]? Alternative Method: