Achieve Week 1, #9

[LaurenAkason2A]

Hello! Could someone help me with #9? I am currently very lost and don't understand the steps to the equation. Thanks!

[chemoyku dis 1B]

Hi!

In the question we are given that Ka = 3.69 and that the acid is placed in a solution with a PH = 2.48.

Since the pH is lower than the Ka, the acid would not dissociate to produce the charged deprotonated species, A- .

To make the PH and pKa equal, the acid would remain as its neutral species instead of producing charged deprotonated species.

I hope this helped.

[Ally Mosher]

First, you need to find the equilibrium constant by doing (.6)^2/(0.3)(0.3), and the answer should be 4. Now that you know the Kc, you make an ICE chart of the new concentrations. The equations should be:

0.3+x, 0.3+x, 0.9-2x

Then you have to solve for x.

Then you plug x into 0.9-2x and that should be your answer.

Hope this helps.

[Phoebe Ko 3E]

To add on to the previous response, the reason why 0.900 M is the initial NO concentration in the ICE chart is because 0.900 M is the NEW concentration after adding NO to the original reaction. Since NO (the product of the original reaction) is added, the reaction now shifts left and favors the reactant N2 and O2. Hence, the product concentration ([NO]) would decrease and the reactant concentrations ([N2], [O2]) would increase to re-establish the equilibrium. Therefore, the concentrations of N2, O2, and NO in the ICE chart at equilibrium would be 0.3+x, 0.3+x, and 0.9-2x, respectively. I hope this was helpful!

[Ginny Ghang 1B]

With the given concentrations, first find the value of Kc. Then, create an ICE chart where the initial value of N2 and O2 are still the given concentrations. The initial concentration of NO will be the new concentration (after more NO is added). Set the value of Kc that you found equal to [NO]^2/([N2][O2]). You'll end up being able to square both sides of the equation as you solve for x. [NO]final will be equal to the NO initial - 2x.

[Symphony Jackson 3K]

For problems like these, I think it is most important depend on the ICE table, I recommend going over lecture or searching online for videos related to problems like these considering how many resources there are about it. For me, the values given in this question could be solved by taking the root of both sides (but it always can be done through a quadratic equation), then I solved for X and input the x values back into the equations!

[Kitana_Garcia_3A]

First, you plug in your initial values and the new concentration for NO that were given into the Kc expression to find the Kc at that time. After that, you plug your initial values into the ICE box and plug in your x's, in order to plug those into another Kc expression that is set equal to the first Kc value that you calculated at the beginning. You can solve this by taking the square root of both sides and then solving for x. Finally, once you get x, you can subtract that from your initial NO concentration and this value is your answer.

[Mahika Saoji 3K]

Hi! First, I found the Kc with the initial concentration of NO which was 0.600 M and I got 4. After that I set up an ICE table, but the key to this step is for the initial concentration of NO we have to use 0.900 because it is the new initial concentration. You then proceed with the usual steps of an ICE table and get these outputs: N2 and O2 are 0.3+X and NO is 0.9-2x. You now set up the values in the Kc equation (P/R) and make theses values equal to 4 (the Kc we previously found). I took the square root of both sides to simplify the equation. You should have 0.9-2X/0.3+X = 2 now. I solved for X and then plugged this value into the ICE equation we got for NO. That's all! I hope that helped!