Acids and Bases 15.5-15.6

15.5 - Autoionization of Water & pH

Water is Amphoteric! Amphoteric: a compound able to react both as an acid and as a base. Even when in a pure form, water acts as an acid and a base with itself. This process is called autoionization. A very small amount of ions will form in pure water. Some molecules of H2O will act as acids, each donating a proton to a corresponding H2O molecule that acts as a base. The proton-donating molecule becomes a hydroxide ion (OH-) while the proton-accepting molecule becomes a hydronium ion (H3O+).

The equilibrium constant is called: the ion product constant for water (kw) (sometimes called the dissociation constant for water) In pure water, since H2O is the only source of these ions, the concentration of H3O+ ions and OH- are equal, and the solution is neutral. Since the concentrations are equal, we easily calculate them from Kw. [H3O+] = [OH-] = √ Kw = 1.0 x 10^-7 (in pure water at 25 degrees C)

Kw = [H3O+] [OH-] = 1.0 x 10^-14 at 25C Formula (on reference table): Kw = [H3O+] [OH-] = 1.0 x 10^-14 at 25C The concentration of H3O+ times the concentration of OH- is always equal to 1.0 x 10^-14 at 25C.

Therefore… If [H3O+] increases, Then [OH-] must decrease in order for the ion product constant (that’s nothing fancy, just kw the equilibrium constant) to remain 1.0 x 10^-14.

In an acidic solution [H3O+] > [OH-] This change would make an acidic solution. An acidic solution contains an acid that creates additional H3O+ ion causing the [H3O-] to increase and the [OH-] to decrease. In an acidic solution [H3O+] > [OH-]

If [H3O+] = 1.0 x 10^-3, what is the concentration of OH-? Walk Through: If [H3O+] = 1.0 x 10^-3, what is the concentration of OH-? Solve the ion product constant expression for [OH-] with the formula from the reference table. Kw = [H3O+] [OH-] = 1.0 x 10^-14 at 25C Is this solution acidic or basic? [H3O+] [OH-] = 1.0 x 10^-14 (1.0 x 10^-3)[OH-] = 1.0 x 10^-14 [OH-] = (1.0 x 10^-14)/(1.0 x 10^-3) [OH] = 1.0 x 10^-11 M Acidic: 1.0 x 10-3 > 1.0 x 10-11 [H3O+] > [OH-]

And… If [OH-] increases, Then [H3O+] must decrease in order for the ion product constant (still just kw the equilibrium constant) to remain 1.0 x 10^-14.

In a basic solution [H3O+] < [OH-] This change would make a basic solution. A basic solution contains a base that creates additional OH- ions causing the [OH-] to increase and the [H3O-] to decrease. In a basic solution [H3O+] < [OH-]

Walk Through If [OH-] = 1.0 x 10^-12, solve the ion product constant expression for [H3O+]. Use the Kw formula from the reference table. Is this solution acidic, basic or neutral? [H3O+] [OH-] = 1.0 x 10^-14 [H3O+] (1.0 x 10^-12) = 1.0 x 10^-14 [H3O+] = (1.0 x 10^-14)/(1.0 x 10^-12) [H3O+] = 1.0 x 10-2 M Basic: 1.0 x 10-2 < 1.0 x 10-12 [H3O-] > [OH-]

Notice: Changing [H3O+] in an aqueous solution produces an inverse change in [OH-] and vise versa.

Summarizing Kw: A neutral solution contains [H3O+] = [OH-] = 1.0 x 10^-7 M (at 25C) An acidic solution contains [H3O+] > [OH-] A basic solution contains [OH-] > [H3O+]

Practice Problem! Calculate [OH-] at 25C for each solution and determine if the solution is acidic, basic, or neutral. [H3O+] = 7.5 x 10-5 M [H3O+] = 1.5 x 10-9 M [H3O+] = 1.0 x 10-7 M

Answers: (7.5 x 10-5 )[OH-] = (1.0 x 10-14) a) [OH-] = 1.3 x 10-10 M, acidic solution b) [OH-] = 6.7 x 10-6 M, basic solution c) [OH-] = 1.0 x 10-7 M, neutral solution (7.5 x 10-5 )[OH-] = (1.0 x 10-14) [OH-] = (1.0 x 10-14 )/(7.5 x 10-5 ) (1.5 x 10-9 )[OH-] = (1.0 x 10-14) [OH-] = (1.0 x 10-14 )/(1.5 x 10-9 ) (1.0 x 10-7)[OH-] = (1.0 x 10-14) [OH-] = (1.0 x 10-14 )/(1.0 x 10-7)

The pH scale: A Way to Quantify Acidity and Basicity pH is defined as the negative of the log of the hydronium ion concentration. Formula: pH = -log[H3O+] Note: write pH to two decimals places because only the numbers to the right of the decimal are significant in a logarithmic A solution with [H3O+] = 1.0 x 10-7 M (neutral at 25C) has a pH of: pH = -log[H3O+] = -log(1.0 x 10-7 ) = 7

In General… pH < 7 Acidic Solution pH > 7 Basic Solution pH = 7 Neutral Solution

Because the pH scale is a logarithmic scale, an increase of 1 on the pH scale corresponds to a factor of 10 decrease in [H3O+]. ACIDIC BASIC

Practice Problems! Calculate the pH of each solution at 25C and indicate whether the solution is acidic or basic. [H3O+] = 1.8 x 10-4 M [OH-] = 1.3 x 10-2 M 2) Calculate the [H3O+] for a solution with a pH of 4.80

Answers 1a) pH = -log[H3O+] = -log(1.8 x 10-4 ) = -(-3.74) = 3.74 (acidic) pH = -log[H3O+] = -log(7.7 x 10-13) = -(-12.11) = 12.11(basic) 1b) First find [H3O+] using Kw. [H3O+] [OH-] = 1.0 x 10-14 [H3O+](1.3 x 10-2 M) = 1.0 x 10-14 [H3O+] = (1.0 x 10-14)/(1.3 x 10-2 M) [H3O+] = 7.7 x 10-13 M

2) pH = -log[H3O+] 4.8 = -log[H3O+] -4.8 = log10[H3O+] 10-4.80 = [H3O+] [H3O+] = 1.6 x 10-5 M Divide by -1 Take the base of the log (base = 10) and raise it to the power of -4.8. Then set that equal to [H3O+] and plug 10-4.80 into your calculator.

pOH & Other p Scales The pOH scale is comparable to the to the pH scale but is defined with respect to [OH-] instead of [H3O+]. It can be found when the [OH-] is known. pOH = -log[OH-] On the pOH scale… pOH < 7 basic pOH > 7 acidic pOH = 7 neutral Ex: A solution with an [OH-] of 1.0 x 10-3 M has a pOH of 3 (basic).

We can derive a relationship between pH and pOH at 25C from the expression for Kw: [H3O+] [OH-] = 1.0 x 10-14 Take the log of both sides: log([H3O+] [OH-]) = log(1.0 x 10-14) log[H3O+] + log[OH-] = -14.00 -log[H3O+] - log[OH-] = 14.00 pH + pOH = 14 Note: the sum of pH and pOH is always 14 at 25C. Therefore, a solution with a pH of 3 has a pOH of 11.

The pKa Scale pKa = -logKa The pKa of a weak acid is another way to quantify its strength. The smaller the pKa , the stronger the acid.

15.6 - Finding the [H3O+] and pH of Strong and Weak Acid Solutions

A solution containing a strong or weak acid has two potential sources of [H3O+]: The ionization of the acid itself HA(aq) + H2O(l) H3O+(aq) + A-(aq) The autoionization of water H2O(l) + H2O(l) H3O+(aq) + OH- The autoionization of water contributes a negligibly small amount of H3O+ compared to the ionization of the strong or weak acid. In a strong or weak acid solution, the additional H3O+ from the acid causes the autoionization equilibrium to shift left. Therefore, the autoionization of water produces even less H3O+ than in pure water and so it can be neglected. We just focus on amount of H3O+ produced by the acid.

Strong Acids Because… Strong acids completely ionize in solution & we can ignore the ions contributed by the autoionization of water The concentration of H3O+ in a strong acid solution is equal to the concentration of the strong acid. Ex: 0.10 M HCl [H3O+] = 0.10 M pH = -log[H3O+] = -log(0.10) = 1.00

Weak Acids The concentration of H3O+ is not equal to the concentration of a weak acid. Calculating the [H3O+] formed by the ionization of a weak acid requires solving an equilibrium problem (Think Chapter 14 ICE Tables). Since we can still ignore the contribution of H3O+ from the autoionization of water, we only have to determine the concentration of H3O+ formed by: HA(aq) + H2O(l) H3O+(aq) + A-(aq)

“The pH of an acid solution depends on both the strength of the acid and the concentration of the acid”(College Board)

Consider a 0. 10 M solution of the generic weak acid HA Consider a 0.10 M solution of the generic weak acid HA. We ignore the autoionization of water and just determine the concentration of H3O+ formed by the following equilibrium: HA(aq) + H2O(l) H3O+(aq) + A-(aq) [HA] [H3O+] [A-] Initial 0.10 ~0.00 0.00 Change -x +x Equilibrium 0.10 - x x

Find the value of x by using the equilibrium expression to set up an equation with x as the only variable. Ka = ([H3O+][A-]) / [HA] Ka= x2 / 0.10 - x (The value of the acid ionization constant will be given for you to substitute in for Ka.)

Practice Problem! Find the [H3O+] of a 0.100 M HCN solution. Ka= 4.9 x 10 -10. [HCN] [H3O+] [CN-] Initial 0.100 Change -x +x Equilibrium 0.100 - x x

Answer Ka = ([H3O+][CN-]) / [HCN] = x2 / (0.100 - x) [H3O+] = 7.0 x 10-6 “x is small” approximation allows you to cross out x and ignore it (7.0 x 10-6 / 0.1) x 100 = 7.0 x 10 -3% Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in approximation. The ratio should be less than 0.05 or 5%. (if x is small is not valid you solve the equation normally with the quadratic equation)

Practice Problem! Find the pH of a 2.00 M HNO2 solution. Ka = 4.6 x 10-4. Hint Think about what value you are missing for the calculation of pH [HNO2] [H3O+] [NO2-] Initial 2.00 Change -x +x Equilibrium 2.00 - x x

Answer Ka = ([H3O+][NO2-]) / [HNO2] = x2 / (2.00 - x) [H3O+] = 3.0 x 10-2 “x is small” approximation allows you to cross out x and ignore it ( 3.0 x 10-2 / 2.00) x 100 = 1.5% Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in approximation. The ratio should be less than 0.05 or 5%.

Answer continued pH = -log[H3O+] = -log[3.0 x 10-2] = =1.52

Practice Problem! (Last one) A 0.100 weak acid (HA) solution has a pH of 4.45. Find the Ka value for the acid. [HA] [H3O+] [A-] I 0.100 C -3.5 x 10-5 +3.5 x 10-5 E 0.100-3.5 x 10-5 3.5 x 10-5 pH = -log[H3O+] 4.45 = -log[H3O+] [H3O+] = 3.5 x 10-5 M

Ka = ([H3O+][A-]) / [HA] = ((3.5 x 10-5)(3.5 x 10-5))/ 0.100 = 1.3 x 10-8

Formulas Recap (On Reference Table) Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25C pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14

Thanks for Listening! (Good Luck on the test!)