Calculate the [NH4+] in solution that is 0.25 M NH3 and 0.20 M NaOH. (Kb for NH3 = 1.8 x 10-5)
I set up ICE from the equation:
NH3 + H20 ------NH4 + OH-
I 0.25 0 0
C -x +x +x
E 0.25-x x x
1.8 *10^-5 = x*x/ 0.25-x
4.5* 10^-6 = x2 took the square root to get:
2.12 * 10^-3 =x
but I got it incorrect so I am wondering what I did wrong in my set up.
Where did you factor in NaOH? The ice chart looks ok. But you must remember that OH^- = x from the NH3 + H2O reaction and 0.2 M from NaOH; therefore, the total OH^- = x+0.2.
Thanks I figured out what I didn't do which was factor in that 0.2M of NAOH
Your initial setup of the ICE table is correct. However, when you set up the expression for the equilibrium constant (Kb) for NH3, you made an error.
Kb = [NH4+][OH-]/[NH3]
Since NH4+ and OH- are both getting produced by the reaction, their concentrations at equilibrium can be assumed to be x. However, the concentration of NH3 at equilibrium is not 0.25-x, but rather its initial concentration of 0.25 M minus the concentration that reacted to produce NH4+ and OH-. In this case, it is (0.25 - x).
Therefore, the correct equation to set up is:
(1.8 x 10^-5) = (x * x)/(0.25 - x)
Solving this equation will give you the correct value for x, and then you can use that value to calculate the concentration of NH4+.
Your initial setup of the ICE table is correct. However, it seems like you made a mistake when solving for x.
To solve for x, you correctly set up the equation Kb = x^2 / (0.25 - x). But when you solved for x, you took the square root of 4.5 x 10^-6, which gave you x = 2.12 x 10^-3. This is where the mistake occurred.
When solving the quadratic equation, you must consider both the positive and negative values of the square root because x can be either positive or negative. In this case, since you are dealing with concentrations (which cannot be negative), the only valid solution is the positive value for x.
So, the correct value for x should be x = 2.12 x 10^-3 M.
To calculate the [NH4+] in solution, you need to consider the stoichiometry of the reaction. For every 1 mol of NH3 that reacts, 1 mol of NH4+ is produced. Therefore, the concentration of NH4+ is also 2.12 x 10^-3 M.
Thus, the [NH4+] in the solution is 2.12 x 10^-3 M.