Learning Objectives Describe the autoionization of water. Calculate the concentrations of H + and OH − in solutions, knowing the other concentration. We have already seen that H 2 O can act as an acid or a base: NH 3 + H 2 O → NH 4 + + OH − (H 2 O acts as an acid) HCl + H 2 O → H 3 O + + Cl − (H 2 O acts as a base) It may not be surprising to learn, then, that within any given sample of water, some H 2 O molecules are acting as acids, and other H 2 O molecules are acting as bases. The chemical equation is as follows: H 2 O + H 2 O → H 3 O + + OH − This occurs only to a very small degree: only about 6 in 10 8 H 2 O molecules are participating in this process, which is called the autoionization of water . At this level, the concentration of both H + (aq) and OH − (aq) in a sample of pure H 2 O is about 1.0 × 10 −7 M. If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have [H + ] = [OH − ] = 1.0 × 10 −7 M for any sample of pure water because H 2 O can act as both an acid and a base. The product of these two concentrations is 1.0 × 10 −14 : [H + ] × [OH − ] = (1.0 × 10 −7 )(1.0 × 10 −7 ) = 1.0 × 10 −14 In acids, the concentration of H + (aq)—[H + ]—is greater than 1.0 × 10 −7 M, while for bases the concentration of OH − (aq)—[OH − ]—is greater than 1.0 × 10 −7 M. However, the product of the two concentrations—[H + ][OH − ]—is always equal to 1.0 × 10 −14 , no matter whether the aqueous solution is an acid, a base, or neutral: [H + ][OH − ] = 1.0 × 10 −14 This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted K w : K w = [H + ][OH − ] = 1.0 × 10 −14 This means that if you know [H + ] for a solution, you can calculate what [OH − ] has to be for the product to equal 1.0 × 10 −14 , or if you know [OH − ], you can calculate [H + ]. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of K w . Example \(\PageIndex{1}\) What is [OH − ] of an aqueous solution if [H + ] is 1.0 × 10 −4 M? Solution Using the expression and known value for K w , K w = [H + ][OH − ] = 1.0 × 10 −14 = (1.0 × 10 −4 )[OH − ] We solve by dividing both sides of the equation by 1.0 × 10 −4 : \[\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber \] It is assumed that the concentration unit is molarity, so [OH − ] is 1.0 × 10 −10 M. Exercise \(\PageIndex{1}\) What is [H + ] of an aqueous solution if [OH − ] is 1.0 × 10 −9 M? Answer 1.0 × 10 −5 M When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H + or OH − ions in the formula unit because [H + ] or [OH − ] may not be the same as the concentration of the acid or base itself. Example \(\PageIndex{2}\) What is [H + ] in a 0.0044 M solution of Ca(OH) 2 ? Solution We begin by determining [OH − ]. The concentration of the solute is 0.0044 M, but because Ca(OH) 2 is a strong base, there are two OH − ions in solution for every formula unit dissolved, so the actual [OH − ] is two times this, or 2 × 0.0044 M = 0.0088 M. Now we can use the K w expression: [H + ][OH − ] = 1.0 × 10 −14 = [H + ](0.0088 M) Divide both sides by 0.0088: \[\left [ H^{+} \right ]=\frac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M\nonumber \] [H + ] has decreased significantly in this basic solution. Exercise \(\PageIndex{2}\) What is [OH − ] in a 0.00032 M solution of H 2 SO 4 ? (Hint: assume both H + ions ionize.) Answer 1.6 × 10 −11 M For strong acids and bases, [H + ] and [OH − ] can be determined directly from the concentration of the acid or base itself because these ions are 100% ionized by definition. However, for weak acids and bases, this is not so. The degree, or percentage, of ionization would need to be known before we can determine [H + ] and [OH − ]. Example \(\PageIndex{3}\) A 0.0788 M solution of HC 2 H 3 O 2 is 3.0% ionized into H + ions and C 2 H 3 O 2 − ions. What are [H + ] and [OH − ] for this solution? Solution Because the acid is only 3.0% ionized, we can determine [H + ] from the concentration of the acid. Recall that 3.0% is 0.030 in decimal form: [H + ] = 0.030 × 0.0788 = 0.00236 M With this [H + ], then [OH − ] can be calculated as follows: \[\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{0.00236}=4.2\times 10^{-12}M\nonumber \] This is about 30 times higher than would be expected for a strong acid of the same concentration. Exercise \(\PageIndex{3}\) A 0.0222 M solution of pyridine (C 5 H 5 N) is 0.44% ionized into pyridinium ions (C 5 H 5 NH + ) and OH − ions. What are [OH − ] and [H + ] for this solution? Answer [OH − ] = 9.77 × 10 −5 M; [H + ] = 1.02 × 10 −10 M Summary In any aqueous solution, the product of [H+] and [OH−] equals \(1.0 \times 10^{−14}\) (at room temperature).