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What are the equilibrium concentrations and pH of these solutions at equilibrium?

40.0 mL of 1.5 M HF with 20.0 mL of 0.500 NaF
0 (0%)
50.0 mL of 0.75 M NaOH with 50.0 mL of 1.00 M HC2H3O2. ka of acetic acid is 1.8x10^-5
0 (0%)

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Topic: Calculating equilibrium concentrations and pH of solutions  (Read 4660 times)

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Offline ewq5555

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Calculating equilibrium concentrations and pH of solutions
« on: February 03, 2011, 04:35:25 PM »
Please calculate the equilibrium concentrations of all ions present in the following solutions. please calculate pH for each solution as well at equilibrium.

a) 40.0 mL of 1.5 M HF with 20.0 mL of 0.500 NaF

b) 50.0 mL of 0.75 M NaOH with 50.0 mL of 1.00 M HC2H3O2.
the ka value of acetic acid is 1.8x10^-5

much thanks. im having trouble with acids and bases so this will be a huge help


Offline rabolisk

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Re: Calculating equilibrium concentrations and pH of solutions
« Reply #1 on: February 03, 2011, 04:52:58 PM »
Show your attempt first and help will be given.

Offline ewq5555

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Re: Calculating equilibrium concentrations and pH of solutions
« Reply #2 on: February 03, 2011, 04:59:37 PM »
ok. im new with this forum so thanks for that advice :D

a) I believe the equation is HF + NaF --> HF + NaF. i calculated that there is 0.06 mol of HF initially and 0.01 mol of NaF initially. so, 0.06-0.01 = 0.05 mol H+ in excess. the total volume is 40mL + 20 mL = 60 mL. so, 0.05/0.060 = 0.83 M = [H+]. however, im not sure this is correct.

b) HA+ OH- --> A-. i set up an equilibrium table to solve for .125 M HC2H3O2 initially and .375 M NaOH initially.

then Ka= [H+][A-]/[HA]. 1.8x10^-5 = x(.375+x)/(.125-x). x=[H+]=5.996x10^-6. pH= 5.22

final answer: [H+]=6.00X10^-6, [C2H3O2-]=.375, pH = 5.22

again, im not sure if i am correct.

Offline rabolisk

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Re: Calculating equilibrium concentrations and pH of solutions
« Reply #3 on: February 03, 2011, 05:22:11 PM »
I'm going to start with the easier one first.

b) This is correct. An easier way of doing this is to use the Henderson-Hasselbach equation. pH = pKa + log([A-]/[HA]). However, your approach is correct, and the two are equivalent. They are only different mathematically.

a) The equation is HF  ::equil:: H+ + F-. The difference between this and (b) is that fluoride is not a strong base that will react with HF all the way. Instead, there is an equilibrium. You will need to use Ka for HF.

Offline ewq5555

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Re: Calculating equilibrium concentrations and pH of solutions
« Reply #4 on: February 03, 2011, 05:37:41 PM »
many thanks rabolisk!

if u see this again, my answers are: [H+]=0.0032 M, [F-]= 0.17 M, and pH=2.5 (u can tell me if im wrong or right haha)

thanks i have a clearer understanding of the topic now.

Offline rabolisk

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Re: Calculating equilibrium concentrations and pH of solutions
« Reply #5 on: February 03, 2011, 06:04:02 PM »
I got a slightly different answer of [H+] = 0.0040M, [F-] = 0.17 M, and pH = 2.4. It's probably due to using different pKa values though. I used 3.17 from wikipedia.

Offline ewq5555

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Re: Calculating equilibrium concentrations and pH of solutions
« Reply #6 on: February 05, 2011, 03:46:12 PM »
yes u are correct. i messed up a step. thanks!

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