Sapling Question 3

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Jaden Ward 3J
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Sapling Question 3

Postby Jaden Ward 3J » Sun Jan 24, 2021 6:04 pm

Muscles produce lactic acid, CH3CH(OH)COOH(aq) , during exercise. Calculate the percent ionization (deprotonation), pH, and pOH of a 0.1024 M solution of lactic acid. The acid‑dissociation (or ionization) constant, Ka , of this acid is 8.40×10−4 .

can someone please help me find the poh, ph, and percent ionization? Thank you.

Alex Benson
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Re: Sapling Question 3

Postby Alex Benson » Sun Jan 24, 2021 6:14 pm

For this question I began by writing out the equation, and from there you can make the ICE table. The ICE table can then give you the equilibrium constant, which in your case should be 8.40*10^-4= x^2/(0.1024-x). From this point you should be able to solve for x using the quadratic formula. Next, to find the percent ionization you would take the x you found and divide that by 0.1024, and then multiply that by 100. The pH can be found by plugging x into -log(x), and then pOH is 14 subtracted by your pH. Hope this helps!

Sameer Chowdhury 3C
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Re: Sapling Question 3

Postby Sameer Chowdhury 3C » Sun Jan 24, 2021 6:17 pm

CH3CH(OH)COOH <--> CH3CH(OH)COO- + H+. This is what the ionization of lactic acid looks like. After knowing this and setting up an ICE table you can find the concentration of H+. Finding that will allow you to find pH and pOH. The percent ionization is the concentration of H+ over the initial concentration of lactic acid times 100.

Levon Grigoryan
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Re: Sapling Question 3

Postby Levon Grigoryan » Sun Jan 24, 2021 6:39 pm

after the quadratic equation you get the value of x. Then you take (Ch3CH(OH)COOH) minus CH3CH(OH)COO-) to find the value of (H+). to find the pH you take the -log) of H+ which gives you the value of the pH. And if you have the pH number, you can easily find the pOH which will be 14 - pH.

Ven Chavez 2K
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Re: Sapling Question 3

Postby Ven Chavez 2K » Sun Jan 24, 2021 6:57 pm

As provided by the other posts, make the ICE table and be meticulous in setting up your Ka equation which will help you find x which will eventually be the Molar concentration of [H+]. To find %Ionization you divide [H+] by the initial concentration of the reactant lactic acid. Then to find pH you find the -log[H+]. To find pOH you can subtract the pH found from 14. Because Ka is 10^-4 I assumed it could be approximated. When approximating, you find, however, that %ionization is higher than 5% which means you cannot approximate so be careful of that.

Anusha Chaudhary 1J
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Re: Sapling Question 3

Postby Anusha Chaudhary 1J » Sun Jan 24, 2021 7:01 pm

I started this question by writing out the equation and making the ICE box. Then, I plugged in the ICE box values into the equilibrium constant and solved. I tried to do the approximation method, but later found that the % ended up being more than 5%, so I had to solve the quadratic equation instead. After that, I found the percent ionization by dividing the x value by the initial lactic acid concentration, and I found the pH by plugging in x into -log(x). Finally, I solved for the pOH by subtracting the pH I found from 14.

Vivian_Le_1L
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Re: Sapling Question 3

Postby Vivian_Le_1L » Sun Jan 24, 2021 7:20 pm

I began by writing out the equation and then made an ice box. I wrote out the Ka expression and solved for x using the quadratic formula. The % ionization is higher than 5%, so you can't approximate for this problem. The value that you find for x is also equal to [H+]. To find the % ionization, divide [H+] by the initial concentration. To find pH, plug [H+] into the equation pH = -log[H+]. To find the pOH, subtract pH from 14.

Anastasia Yulo 1C
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Re: Sapling Question 3

Postby Anastasia Yulo 1C » Sun Jan 24, 2021 8:02 pm

It helps to remember these formulas:
-log(Ka)=pKa, -log(Kb)= pKb

pKa + pKb = 14
Ka x Kb =10^-14

and pH+pOH=14

Nan_Guan_1L
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Re: Sapling Question 3

Postby Nan_Guan_1L » Sun Jan 24, 2021 8:29 pm

just to add on to the discussion,
when i did my calculations, i remember double checking to see if we can use approximation. but since Ka of this acid is 8.40×10−4, I remember I had to solve the quadratic equation to get the accurate answer. This might be the reason you didn't get the right answer. if Ka is less than 10-4, then usually we can use approximations.


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