Question

I am going to a lab next week on Acids, Bases, PH and Buffers, but i need some help, especially on Buffers.

Q.5 Which solution(s) showed the greatest and which solutions(s) showed little or no change change in pH? Why? regarding which ones, i do understand that i need to do the lab first.

I understand bufferss resist a change in PH when a small amount of strong acid OR strong base is added to it. This means that the PH of the buffers does not change significantly (by more than 1 PH unit) when a small amount of strong acid or base its added Can you please give me an example? i dont understand when say (by more than 1 unit)

In contrast, unbuffered solutions will show significant PH changes ( greater than 1 PH unit) when a small amounts of strong acid or base are added. Can you give an example? i dont understand where it say (greater than 1 PH unit) i will appreciate your help.

Answer 1

Let us take a buffer of CH₃CO₂H and CH₃CO₂Na.

Now calculate pH of a solution prepared by mixing 100.0 mL of 0.10 M CH₃CO₂H (Ka=1.8 x 10^-5) with 25.00 ml of 0.20 M CH3CO2Na. And then adding calculate change in pH when 25.00 mL of 0.10 M NaOH is added to the buffer.

First we will determine number of moles (or mmoles) of the acid and conjugate base:

100 mL x 0.10 M = 10 mmol CH₃COOH
25.00 mL x 0.20 M = 5 mmol CH₃COONa

Total volume of the solution = 25 mL + 100 mL = 125 mL.

to find the new concentration, but leaving it in mmoles is fine, too (for the most part, sometimes the H+ or OH- is significant enough to affect it, but not in this case) When you add CH₃COONa, you're adding the conjugate base to the solution. So you would start out in your initial concentration 5 mmol of CH₃COONa.

CH₃COOH      CH₃COO^- + H⁺
IC: 10 5 0
C -x +x +x
EC 10-x 5+x x

Ka = [CH₃COO^-] [H⁺] / [CH₃COOH]

1.8 x 10^-5 = (5+x) x / (10-x)

Let's make the assumption that x is going to be very small. So:
5+x = ~5; 10-x = ~10

1.8 x 10^-5 = 0.5 x

x = [H⁺] = 3.6 x 10^-5;

pH = -log [H⁺] = -log (3.6 x 10^-5) = 4.44

When we add 25.00 mL of a 0.10 M NaOH solution, it will react completely with CH₃COOH.

NaOH will release 25.00 x 0.10 = 2.5 mmol OH^- ions, which will react completely with CH₃COOH. This will add to the concentration of CH₃COO^- already in solution.

....CH₃COOH + NaOH ----> CH₃COONa + H₂O
I........10.............2.5..... ..........5....................
C.......-2.5..........-2.5.... .........+2.5.................
E.......7.5.............0..... ............7.5.................

Of course, now CH₃COOH can do its thing and disassociate:

CH₃COOH    CH₃COO^- + H⁺
IC 7.5 7.5 0
C -x +x +x
EC 7.5-x 7.5+x x

Ka = [CH₃COO^-][H⁺] / [CH₃COOH]

1.8 x 10^-5 = (7.5+x) x / (7.5-x)

(We'll again make the assumption that x is going to be very small):

So,

1.8 x 10^-5 = 7.5 x / 7.5

x = 1.8 x 10^-5 = [H⁺]

pH =  -log [H⁺] = -log (1.8 x 10^-5) = 4.74

Change in pH after addition of base = 4.74 - 4.44 = 0.30 (which is less than 1).

Let us calculate the pH change caused by adding 10.0 mL of 3.13 M NaOH to 240.0 mL of pure water. This is a case of unbuffered solution.

Pure water : pH = 7.00

Number of moles = volume(L) x molarity

so, moles NaOH added = (0.01 L x 3.13) = 0.0313 moles

[OH^-]= 0.0313 mol / 0.240 L= 0.13 M

pOH = -log[OH^-] = - log(0.13) = 0.88

pH = 14 - pOH = 14 - 0.88 = 13.12

So, change in pH = 13.12 - 7 = 6.12 ( which is greater than 1)