Well, the problem is that one-step solutions are usually shortcuts to be used only AFTER you understand doing it the long way (making an ICE table). I'll try my best though to help you understand the concepts behind the one-step shortcut while keeping my solution to just one-step."If 10 mL of 20M CH3COONa and 10 Ml of .10 M CH3COOH (Ka= 1.7 X 10 to the -5) are mixed, what is a one step solution that can get me the value of pH in this solution?"
CH3COOH + H2O
CH3COO- + H3O+
Ka = ([CH3COO-][H30]) / [CH3COOH]
-log(Ka) = -log (([CH3COO-][H3O+]) / [CH3COOH])
pKa = -log(H3O) + -log([CH3COO-]/[CH3COOH])
pKa = pH - log([CH3COO-]/[CH3COOH])
pH = pKa + log([CH3COO-]/[CH3COOH])
The equation I just derived for you is called the Henderson-Hasselbach equation. It basically allows you to calculate the pH of an acid buffer solution (an acid buffer is a solution of a weak acid and its conjugate base.). The general form for the Henderson-Hasselbach equation for a generic monoprotic acid that dissociates: HA + H2O
A- + H3O+
pH = pKa + log([A-]/[HA]) H-H equation
Since your solution is an acid buffer, all you have to do is plug in the numbers.
pH = pKa + log (20/0.10) = 7.07
"Indicators change colors about 1-1.5 pH units. For bromthymol blue, the pKa is about 7.0. Estimate the pH range over which the color change occurs for bromcresol green indicator."
4-5.5
"What is the pH of a solution prepared by mixing 2.0 mL of a solution of a strong acid having a pH= 2.0 with a 4.0 mL of a similar solution of pH =4.0?"
pH = -log[H3O+] = 2.0
[H3O+] = 10^-2
pH = -log[H3O+] = 4.0
[H3O+] = 10^-4
mol H3O+ = [H3O+]V + [H3O+]V = (10-2M)(.002L) + (10^-4M)(.004L) = 0.0000204 mol H3O+
[H3O+] = mol H3O+/ L H3O+ = 0.0000204 mol / (0.002 + 0.004) = 0.0034 M
pH = -log[H3O+] = -log(0.0034M) = 2.47
"Suppose one of the solutions used in the prior question was a buffer solution instead of a strong acid. Would the resulting pH be the same as above? Why?"
No it would not. A buffer is defined as a solution that resists changes to its pH through its acid/base conjugate. The resulting pH after mixing would therefore be closer to the pH of the buffer (as long as the buffer's buffering capacity hadn't been exceeded).
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