Achieve 2, #8

[Natalie Flores 1F]

Hello, I'm having trouble solving #8 for this week's homework. I don't really know how to begin solving it or what step to take first. Is this problem just like any other one where we go straight into the ICE chart or is there a step I need to do prior to that? If anyone could help me out, I'd really appreciate it!

[Sanjana Sivakumar 2E]

Hello!

This problem is just like the others, however there is one small step you need to do first. Here, you are given the Kb value, but we need the Ka. So, to convert Kb to Ka, you have to divide 10^-14 by Kb. This is because Ka x Kb = 10^-14.

After this, it's pretty much like the rest of the assignments.

Hope this helps!

[Furkan Acar 3C]

As stated before, you need to find Ka as we already have Kb. Since someone else already answered I won't go into great detail but life is easy after finding Ka.

[Amine Adlouni 1L]

Because the question is asking for pH, first use the Kb provided to find the Ka. Then proceed as normal, using the ICE table for the newly calculated Ka.

[305597516]

Hello! In question #8, you are first given Kb, but asked to find the pH of an acid. Therefore, you must convert Kb to Ka using the formula Ka x Kb = Kw. You can do this because the weak acid and base given are conjugates of each other. You can then make an ICE table (although I guess you can make that first and then do the Ka Kb stuff) using the given molarity of the weak acid. Assuming that the acid is simply dissociating, I would put the weak acid they gave you as the reactant (e.g. if NH4Cl's pH is what they want you to find, then NH4+Cl --> NH3 + H+, although Cl is kinda irrelevant and doesn't need to be put here). From the results of your ICE table, you can combine this with the Ka in order to find the value of x, which also serves as the value of the products (they should be equal to each other, supported by the fact the reactant-product ratio is 1:1). You can find the pH by taking the concentration of H+ (H3O+ is like the same thing) and inputting it into -log[H+]. Hope this helped!

[Milli Zunich 3I]

So we still need to use ICE, correct?

[Raymond_Kallely_1A]

Hi, so the first step would be finding the value of Ka. You can find this by using what's given, specifically that the value of Kb is 1.8*10^-5. Since we know that Ka*Kb=1.0*10^-14. After solving for Ka, you can set up an ice table using what is given about the dissociation of this ammine, you can make an ICE table where x^2/0.041-x. After solving for x, all that's left to do is to use the equation, [NH4+]=-log(x). With this, you should get 5.3. Hope this helps

[Daniela G 2C]

Yes, there is one step prior to the ICE chart. I first found out Ka by using the given Kb. I used the equation Ka * Kb = 1.0*10^-14. Once you find Ka, use the ICE chart to find x/[H+]. From the ice chart, our x is equal to H+. Once you find x, you can plug in x to find pH using the formula ph = -log [H+].