Weak acid dissociation problem. Hi, I am stuck on a general chemistry lab and wo

Question

Weak acid dissociation problem.

Hi, I am stuck on a general chemistry lab and would like some help. I need to find the equilibrium concentration of [H+], [X-], [HX], and Ka for each part. the temperature is never 25C, and this is an weak acid dissociation. After that we have to do it again but for HY this time, but the instructions are the same. So you probably do not need to help me with that one.

Part 1 is just measuring the pH of a 10mL sample of .995M of Hx.

I found the pH to be 2.28 with that I found the equilibrium concentration of H+ to be .3579M the X- is .3579M, and the [HX] is .995M.

Part2. Add 20mL of 1M NaA to 10 ml of 1M HA. Mix thoroughly and take 10 mL of the solution and measure its pH. Save the remaining buffer solution for part 3.

For this one I know the initial concentration is just taken from the ones in part1. The pH I found is 4.97. I need to figure out how to find the dilute concertation of [X-] and [HX], and the three equilibrium concentration for [HX], [X-], [H3O+], and the Ka. I am not sure how to find any of the five things other than the equilibrium concentration of [X-] and [H3O+]. Which I just need to take the -log(pH).

Part 3. Dilute 10 ml of the solution for part 2 with 50 mL distilled water. stir and measure the pH of this solution.

For this one I need to find the concentration of [HX] and [X-] after dilution with water as well as the three equilibrium concentration for [HX], [X-], [H3O+], and the Ka. Once again I am completely lost on how to do any of these other than the equilibrium concentration of [X-] and [H3O+]. which is just the -log(pH).The pH for this one is 4.99.

Part 4. mix 10 mL of 1.05M NaOH with 25 mL of HA, stir and record the pH of this solution.

For this one I need to find the concentration of unrelated [HX], concentration of [X-] after reaction, the three equilibrium concentration of [HX], [X-], [H3O+], and the Ka. The pH for this one is 4.47.

Finally there are two questions at the end of the lab.

1. Explain why Ka from part 1 is excluded from the average Ka calculation.

I think this is because I change from NaX to NaOH, but please correct me if I am wrong.

2. Explain why the approximation method can be used for the calculation of Ka for HX but not for HY.

I really don't unerstand this question.

Please help, I really do not understand this lab.

The Ka is not supposed to be given. Next to it, the lab manual states "show work" so I have to figure it out myself. I think I can through the Henderson Hasslebalch equation, but to do so I have to figure out the dilute concentraion of  [X-] and [HX] first. Then with the HHE I can find the Ka, and take the antilog to find the [H3O+].

Explanation / Answer

1)

Part 1 is just measuring the pH of a 10mL sample of .995M of Hx.

Let the acid dissociate as

HX -> H+ + X-

The pH = 2.28

so [H+] = 0.0052 = [x-] = equilibrium concentrations

Initial concentration of HX = 0.995

Equilibrium concentration = 0.995 - 0.0052

Ka = [H+] [x-] / [HX]

Ka = 0.052 x 0.0052 / 0.995 - 0.0052

0.00052 << 0.995

Ka = 2.72 X 10^-5

2) Add 20mL of 1M NaA to 10 ml of 1M HA. Mix thoroughly and take 10 mL of the solution and measure its pH. Save the remaining buffer solution for part 3.

In this part the acid is different from first part ( HA instead of HX). so how can you take concentration from part 1?

The pH you found is 4.97.

From Hendersen equaiton

pH = pKa + log [salt] / [acid]

pH = pKa + log 1

pKa = pH = 4.97

Ka = 1.07 X 10^-5

[A-] = Mainly due to salt

Moles of A- = 1 X 0.02

so concentration = 0.02 / Total volume = 0.02 / 0.03 = 0.667 M

[H+] = 1.07 X 10^-5

PArt 3 :

total volume of above solution = 30mL

we used = 10 ml of the solution for part 2

Diluted with = 50 mL distilled water.

pH = 4.99

so [H+] = 1.023 X 10^-5

[X-] = Moles of X- in 10 mL of solution / total volume = 10 X 0.667 / 50 = 0.1334 M

Part 4. mix 10 mL of 1.05M NaOH with 25 mL of HA, stir and record the pH of this solution.

pH = 4.47 = pKa + log [salt] / [acid]

NaOH + HA --> NaA + H2O

Moles of NaOH = moalrity X volume = 1.05 X 0.025 = 0.0262 moles

So moles of acid neturalized = 0.0262

so moles of salt formed = 0.0262

The pH for this one is 4.47.

pH = pKa + log [salt] / [acid]

pKa = from part 3 = 4.97

4.47 = 4.97 + log [0.0262] / [acid]

-0.5 = log [0.0262] / [acid]

Taking antilog

[acid] = 0.0262 / 0.316 = 0.0829 M

Answers to question

1) Yes you are right

2) the approximation method can be used for the calculation of Ka for HX as Hx is a weak acid

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