What is the original molarity of a solution of ammonia whose pH is 11.11

okay i tried to convert to pOH- so i got 2.89 and then to [OH]=1.288e-3

and then solve using Kb=1.8e-5

whats going wrong???

The pOH is correct as is the (OH)^-. Show your work for the rest of it.

I would do
[(1.288 x 10^-3)^2/(y-1.288 x 10^-3)] = 1.8 x 10^-5 and solve for y. I get 0.0935 M but check my work. Without the correction for y-0.001288 (that is, calling it just y), I get 0.0922 M

Well, it seems like your calculations took a detour and ended up in a confusing neighborhood. Let's find the right direction, shall we?

To determine the original molarity of a solution, you need to start by converting the pH to pOH correctly. Since pH + pOH = 14, if the pH is 11.11, the pOH is 14 - 11.11, which is 2.89. Well done so far!

Now, to find the hydroxide ion concentration ([OH-]), you cannot just use a random number out of thin air (or lab accidents, for that matter). Instead, you need to find it using the relationship [OH-] = 10^(-pOH). In this case, [OH-] = 10^(-2.89), which turns out to be approximately 1.28 x 10^(-3) M. Great job!

Finally, you can use the equilibrium constant expression for the base dissociation reaction (Kb) to determine the original molarity (M) of the solution:

Kb = [OH-]^2 / [NH3]

Plugging in the values:

1.8 x 10^(-5) = (1.28 x 10^(-3))^2 / [NH3]

Now it's just a matter of solving for [NH3] algebraically. Give it a shot, and if you need any more puns or jokes along the way, I'm here to provide amusement!

Based on the information you provided, it seems like you are attempting to calculate the original molarity of an ammonia solution, given its pH and the concentration of hydroxide ions. However, there seems to be a mistake in your calculations.

To find the concentration of hydroxide ions ([OH-]), you correctly converted the pH to pOH and obtained 2.89. However, when converting pOH to [OH-], the correct equation is [OH-] = 10^(-pOH). Therefore, [OH-] should be equal to 10^(-2.89), which is approximately 1.292 x 10^(-3) M, not 1.288 x 10^(-3) M.

Using the equation for the base dissociation constant of ammonia (Kb = [OH-][NH4+]/[NH3]), you can rearrange it to solve for [NH3] (original molarity of ammonia). Since [OH-] is known as 1.292 x 10^(-3) M and Kb is 1.8 x 10^(-5), you can substitute these values into the equation to find [NH3].

Kb = [OH-][NH4+]/[NH3]
1.8 x 10^(-5) = (1.292 x 10^(-3))^2 / [NH3]

Solving for [NH3], you would divide (1.292 x 10^(-3))^2 by 1.8 x 10^(-5).

To find the original molarity of a solution of ammonia, you need to calculate the concentration of hydroxide ions ([OH-]) first.

Given that the pH of the solution is 11.11, it indicates a basic solution. Since pH + pOH = 14, you can calculate the pOH as follows:

pOH = 14 - pH = 14 - 11.11 = 2.89

Now, in order to convert pOH to [OH-], you need to take the antilog of the pOH value:

[OH-] = 10^(-pOH) = 10^(-2.89) = 1.29 x 10^(-3) M

So far, your calculations are correct.

To find the original molarity (concentration) of the ammonia solution, you can use the relationship between the concentration of hydroxide ions ([OH-]) and the base dissociation constant (Kb) for ammonia.

Kb = [NH4+][OH-] / [NH3]

Given that Kb = 1.8 x 10^(-5) and [OH-] = 1.29 x 10^(-3), you can rearrange the equation to solve for the concentration of ammonia ([NH3]):

[NH3] = [OH-] * Kb / [NH4+]

Here's where the issue lies. To proceed, we need the concentration of ammonium ions ([NH4+]), which was not provided in the question. Without that information, it is impossible to calculate the original molarity of the ammonia solution.

Therefore, it seems that you are not doing anything wrong in your calculations. It's just that the missing concentration of [NH4+] prevents you from finding the original molarity of the ammonia solution.