16.2 Buffers

Buffers: What are they? A buffer is a substance that can resist the change in pH by neutralizing added acid or base. – A buffer contains: 1.) significant amounts of a weak acid AND a conjugate base. 2.) significant amounts of a weak base AND its conjugate acid. – In either cases, they must contain both!!! The way a buffer works: – Let’s take H 2 CO 3 and it’s conjugate base HCO 3 -. If base is added to the buffer, the weak acid reacts with the base, neutralizing it! If acid is added to this buffer, the conjugate base reacts with the acid, neutralizing it! Again, both must be present (and will only work if the amount of acid or base added is less than the amount of weak acid or the amount of the conjugate base.

Let’s Try a Practice Problem! Which solution is a buffer? (a)A solution that is M in HNO 2 and a M HCl (b)A solution that is M in HNO 3 and a M NaNO 3 (c)A solution that is M in HNO 2 and a M NaCl (d)A solution that is M in HNO 2 and a M NaNO 2 (d) A solution that is M in HNO 2 and a M NaNO 2 (This is because nitrous acid is a weak acid, and the sodium ion is neutral, but the nitrite ion is the conjugate base of nitrous acid.)

Calculating the pH of a Buffer Solution When calculating the pH of a buffer solution that contains common ions, we work an equilibrium problem in which the initial concentrations include both the acid and its conjugate base. – For example, if we had to calculate the pH of a buffer that is made up from HC 2 H 3 O 2 and NaC 2 H 3 O 2. (we will work an example problem out on the following slide). In there are common ions, the ion (for example, the conjugate base of an acid) causes the acid to ionize even less. This is because the presence of that ion (being that in is a product) shifts the equilibrium left. This is known as the common ion effect.

Let’s Try a Practice Problem! Calculate the pH of a buffer solution that is M in HC 2 H 3 O 2 and M in NaC 2 H 3 O 2. (The K a value of acetic acid from table 15.5 in your text is: 1.8x10 -5 ) First, let’s write the reaction showing the ionization of the acid in water. HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) Next, we need to set up an ice table, and we can plug in an initial concentration of both acetic acid, as well as the acetate ion. [H 3 O + ][C 2 H 3 O 2 - ] K a = continued  [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ] M[H 3 O + ] M[C 2 H 3 O 2 - ] M Initial0.100~ Change-x+x Equilibrium xx x

[H 3 O + ][C 2 H 3 O 2 - ] K a = [HC 2 H 3 O 2 ] (x)(0.100+x) 1.8x10 -5 = x 0.100x = 1.8x10 -6 X = 1.8x10 -5 Now check if x is valid. (1.8x10 -5 / 0.100) x 100 = % < 5% so x is valid. x = H 3 O + = 1.8x10 -5 M pH = -log(1.8x10 -5 ) = 4.74

The Henderson-Hasselbalch Equation This equation can be used to simplify the calculation of the pH of a buffer solution. Let’s look at the generic reaction of a weak acid ionizing: HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) We can use the Henderson-Hasselbalch equation to calculate the pH of a buffer in a simple step (as long as the x is small approximation is valid). The H-H equation is listed on your reference table, and is: [A - ] pH = pKa + log  If you use this method, after you calculate the pH, [HA] convert it to the hydronium ion concentration to check if x is valid.

Let’s try a Practice Problem! Calculate the pH of a buffer solution that is M in HCN and M in KCN. For HCN, K a = 4.9x (pKa = 9.31). I will use the H-H approach. [A - ] (0.170) pH = pK a + log = log = 9.14 [HA] (0.250) pH = -log[H 3 O + ] 9.14 = -log[H 3 O + ] [H 3 O + ] = = 7.24X So… (7.24x /0.250) x100 = 2.90x10 -7 % < 5% so x is valid!!!

Let’s Try Another!!! A buffer contains a weak acid HA and its conjugate base A -. The weak acid has a pK a of 4.82 and the buffer has a pH of Which statement is true of the relative concentrations of the weak acid and conjugate base of the buffer? (a)[HA] > [A - ] (b)[HA] < [A - ] (c)[HA] = [A - ] (a) [HA] > [A - ] Which buffer component would you add to change the pH of the buffer to 4.72? We could have to add more of the base [A - ]

Calculating pH Changes in Buffer Solutions Buffers resist changes in pH when either an acid or base is added…however, the pH does change a little. In order to calculate this change, we must take two things into consideration: – 1.) The stoichiometry calculation – 2.) The equilibrium calculation Remember, just as calculated in a problem we did at the beginning of this PPT, of the concentrations of the weak acid and conjugate base are equal, so are the pH and pK a values. From the college board: (So the next few slides are beyond the scope of the AP exam!)

Let’s Try a Practice Problem! (I will walk you through this one.) A 1.0 L buffer solution contains mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2. The value of K a for HC 2 H 3 O 2 is 1.8x Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pK a = -log(1.8x10 -5 ) = Calculate the new pH after adding mol of solid NaOH to the buffer. For comparison, calculate the pH after adding mol of solid NaOH to 1.0 L of pure water. (Ignore any small changes in volume that might occur upon addition of the base). Okay, so first we must use the stoichiometry to show how the addition of a strong base is neutralizing our acid. OH - (aq) + HC 2 H 3 O 2 (aq)  H 2 O(l) + C 2 H 3 O 2 - (aq) On the next slide, we will set up a table to show how the addition of the strong base will affect the weak acid and conjugate base. (This is not an ICE table)

OH - (aq)HC 2 H 3 O 2 (aq)C 2 H 3 O 2 - (aq) Before addition~ mol Addition mol mol mol After addition0.010 mol0.090 mol0.110 mol OH - (aq) + HC 2 H 3 O 2 (aq)  H 2 O(l) + C 2 H 3 O 2 - (aq) Now, that we can see how the addition of a strong base affected our acid (lower the amount) and our conjugate base (increasing the amount), we can set up an ICE tablewith the amounts found above to show the ionization of the weak acid, and find the pH of the buffer after the addition of NaOH. HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) [H 3 O + ][C 2 H 3 O 2 - ] (x)(0.110+x) K a = x10 -5 = continued  [HC 2 H 3 O 2 ] x [HC 2 H 3 O 2 ] M[H 3 O + ] M[C 2 H 3 O 2 - ] M Initial0.090~ Change-x+x Equilibrium xx x

(x)(0.110+x) 1.8x10 -5 = x 0.110x = 1.62x10 -6 x = 1.47x10 -5 (1.47x10 -5 / 0.090) x 100 = 1.63x10 -4 % Since x = H 3 O + = 1.47x10 -5 M pH = -log (1.47x10 -5 ) = 4.83 Now let’s say that instead of setting up the ICE table, I used the H-H equation to solve for the pH of the buffer…it would still work because the x is small approximation is valid. Watch: [C 2 H 3 O 2 - ] (0.110) pH = pKa + log = log = 4.83 [HC 2 H 3 O 2 ] (0.090)

Let’s Try Another!!! A buffer contains equal amounts of a weak acid and its conjugate base and has a pH of Which would be a reasonable value of buffer pH after the addition of a small amount of an acid. (a)4.15 (b)5.15 (c)5.35 (d)6.35 (b) 5.15, since we are dealing with a buffer, the addition of an acid would slightly decrease the pH of the buffer.

NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) [NH 3 ] (0.47) pH = pKa + log = ? + log = [NH 4 + ] (0.23) We need to calculate pKa, and we know that the pKb for ammonia is So, pKa + pKb = 14. pKa = 14 – 4.75 = 9.25 [NH 3 ] (0.47) pH = pKa + log = log = 9.56 [NH 4 + ] (0.23) Check if x is small…. [H 3 O + ] = = 2.75x I can already tell that x will be valid.

Buffers Containing a Base and its Conjugate Acid In order to calculate the pH of a buffer that contains a base and its conjugate acid, we would take the same steps we did previously. However, if we wanted to use the H-H equation, we would need to calculate pK a. Let’s Try a Practice Problem! Calculate the pH of 1.0 L of a buffer solution that is 0.50 M in NH 3 and 0.20 M in NH 4 Cl, upon the addition of 30 mL of 1.0 M HCl. For ammonia, the pK b = Let’s check the stoichiometry and see how the addition of acid affects the amount of weak base and its conjugate acid. H + (aq) + NH 3 (aq)  NH 4 + Now, we could attempt to use the H-H equation to calculate the pH of the buffer after the addition of HCl. continued  H + (aq)NH 3 (aq)NH 4 + (aq) Before addition00.50 mol0.20 mol Addition+0.03 mol-0.03 mol+0.03 mol After Addition0.03 mol0.47 mol0.23 mol

16.2 pgs #’s 28, 30 a, 38, 42 a, 46 Read pgs