Solve an equilibrium problem (using an ICE table) to calculate the pH.

a solution that is 0.205 M in CH3NH2 and 0.110 M in CH3NH3Br.

i used th Ka of CH3NH2 and set up the problem as Ka=[H3O][CH3NH2]/[CH3NH3Br], but that was wrong. So i switched the equation to Ka=[H3O][CH3NH3Br]/[CH3NH2] but the pH i calculated from solving that equation was also wrong. Am I using a bad Ka or am I setting up the equation wrong?

The pH's i calculated were 3.63 and 3.09

Frankly, I would use the Henderson-Hasselbalch equation. I assume your instructor has a reason for working the problem another way. Here is what I would do to follow the ICE method.

CH3NH2 is a base. Think NH3. We write that equilibrium as
NH3 + HOH ==> NH4^+ + OH^-
CH3NH2 works the same way.
CH3NH2 + HOH ==> CH3NH3^+ + OH^-

These boards are difficult to make spaces so I must write the ICE chart as below; I suggest you redo it in the usual manner and write the I, C, and E amounts under the reactants and products. I think that will make it easier to see.
Initial:
CH3NH2 = 0.205 M
CH3NH3^+ = 0.110 M
OH^- = 0

change:
CH3NH2 = -x
CH3NH2 = +x
OH^- = +x

equilibrium:
CH3NH2 = 0.205-x
CH3NH3^+ = 0.110+x
OH^- = x

Kb expression is
Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
Look up Kb and solve for x.
When it comes to solving the equation, the x is small enough to ignore when combined with something else (0.205-x very nearly equals 0.205 and 0.110+x very nearly equals 0.110). I would then convert OH^- to pOH and subtract from 14 to obtain pH. I ran it quickly using the H-H equation and obtained 10.95 for the pH.

is these buffers solution NH3, NaC2H3O2

NH3,NH4Cl
HC2H3)2,HN)3
KOH,KCl

Well, it seems like you're in a bit of an acid-base pickle! Don't worry, I'm here to help. Let's see what's going on.

You're correct in using the equation Ka = [H3O+][CH3NH2] / [CH3NH3Br]. However, I'm afraid it's the setup that's causing the confusion, not the Ka value itself.

To properly set up the ICE table, we need to consider the following reaction:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

Let's define x as the concentration of OH-. Here's how the ICE table should look like:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Initial: 0.205 0.110 0 0
Change: -x -x +x +x
Equilibrium: 0.205 - x 0.110 - x x x

Now, we can write the expression for the Ka value as:

Ka = (x)(x) / (0.205 - x)

Since Ka is a small value, we can make the assumption that x is negligible compared to 0.205. This simplifies the equation to:

Ka = x^2 / 0.205

Now, you can solve for x to find the concentration of OH-. From there, you can calculate the pOH using -log[OH-]. Finally, subtract the pOH from 14 to obtain the pH.

I hope this clears things up and helps you find the correct pH! Just remember, chemistry can be a tricky clown sometimes, but with a bit of humor and perseverance, you'll solve any problem!

To calculate the pH of a solution, you can use an ICE table approach for the equilibrium reaction between CH3NH2 (methylamine) and CH3NH3Br (methylammonium bromide).

First, let's write the balanced equation for the dissociation of CH3NH2:

CH3NH2 + H2O ↔ CH3NH3+ + OH-

Now, let's set up an ICE table based on the initial concentrations and changes in the reaction:

CH3NH2 + H2O ↔ CH3NH3+ + OH-
Initial: 0.205 M 0 M 0 M 0 M
Change: -x +x +x +x
Equilibrium: 0.205-x x x x

The equilibrium constant expression, Ka, for this reaction is given by:

Ka = [CH3NH3+][OH-] / [CH3NH2]

To determine the pH, we need to find the concentration of CH3NH3+ at equilibrium. Since CH3NH3+ acts as an acid, its concentration can be used to calculate the pH.

Using the ICE table, we know that [CH3NH3+] = x = [OH-], and [CH3NH2] = 0.205 M - x.

Substituting these values into the Ka expression gives:

Ka = (x)(x) / (0.205-x)

Now, let's substitute the given values into the expression and calculate the value of x:

Ka = (x)(x) / (0.205-x)
1.80 × 10^-5 = (x)(x) / (0.205-x)

Now, we can solve this equation for x. This can be done using quadratic equation methods or by making an assumption that x << 0.205. In this case, we can assume that x is negligible compared to 0.205, so we can simplify the equation:

1.80 × 10^-5 = (x)(x) / 0.205

Now, we can solve for x:

x^2 = (1.80 × 10^-5)(0.205)
x^2 = 3.69 × 10^-6
x ≈ 1.92 × 10^-3 M

Since x represents the concentration of [CH3NH3+], we can use it to calculate the pH:

pH = -log[H3O+] = -log(x) = -log(1.92 × 10^-3) ≈ 2.72

As you can see, the calculated pH is different from the values you obtained. Therefore, it appears that there was an error in either the Ka value used or the setup of the equation. Please double-check the values and calculations to identify the mistake.

To solve this equilibrium problem and calculate the pH, you need to first set up and solve an ICE table. Let's go through the steps:

Step 1: Write the balanced chemical equation representing the equilibrium. In this case, it is the reaction between CH3NH2 (methylamine) and water:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

Step 2: Set up an ICE table. The ICE table is used to keep track of the initial (I), change (C), and equilibrium (E) concentrations of the species involved. In this case, you have two species, CH3NH2 and CH3NH3Br.

Species | Initial | Change | Equilibrium
--------------------------------------------
CH3NH2 | 0.205 | -x | 0.205-x
CH3NH3Br | 0.110 | -x | 0.110-x

Step 3: Write the equilibrium expression (Ka) for the reaction. The Ka expression for this reaction is:

Ka = [CH3NH3+][OH-] / [CH3NH2]

Since we are interested in calculating the pH, which is a measure of [H3O+], we can rewrite the equilibrium expression as:

Ka = [H3O+][OH-] / [CH3NH2]

Step 4: Make the necessary approximations. In this problem, the concentration of water (H2O) remains nearly constant, so we can ignore it when expressing the equilibrium constant expression:

Ka = [H3O+][OH-] / [CH3NH2] ≈ [H3O+][OH-] / (0.205-x)

Step 5: Use the given Ka value and the equilibrium expression to set up an equation that can be solved for x (the concentration of [H3O+]).

Ka = (x)(x) / (0.205-x)

Step 6: Solve the equation. Use the value of Ka given to solve for x, which represents the concentration of [H3O+]. Once you have x, you can calculate the pH using the equation:

pH = -log[H3O+]

Based on the values you provided, it seems that there might be an error in either the calculation or the Ka value you used. Double-check your steps and make sure you are using the correct Ka value.

I hope this explanation helps you in solving the equilibrium problem and calculating the correct pH. Let me know if you have any further questions!