Calculate the [CH3NH3+] of a 7.02×10-3 M solution of the weak base CH3NH2 (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration). Round your answer to 3 significant digits.
CH3NH2 + H2O = CH3NH3+ + OH-
Set up an ICE chart and substitute into the Kb expression.
Kb = ?? = (CH3NH3^+)(OH^-(/(CH3NH2)
(CH3NH3^+) = (OH^-) = x
((CGH3NH2) = 7.02 x 10^-3
To calculate the concentration of [CH3NH3+], we need to use the given equilibrium equation:
CH3NH2 + H2O = CH3NH3+ + OH-
We can assume that the initial concentration of CH3NH3+ is negligible compared to the concentration of CH3NH2, which means that we can approximate the concentration of CH3NH3+ as zero initially.
Let's denote the equilibrium concentration of CH3NH3+ as x.
The concentration of OH- is equal to the concentration of CH3NH3+ since both are produced in a 1:1 stoichiometric ratio. Therefore, [OH-] = x.
Since the equation shows that one CH3NH2 molecule produces one CH3NH3+ and one OH-, the concentration of CH3NH2 decreases by x, and the concentration of H2O decreases by x.
Therefore, [CH3NH2] = 7.02×10^-3 M - x, and [H2O] = 7.02×10^-3 M - x.
We can now write the equilibrium expression for the reaction:
Kb = ([CH3NH3+][OH-])/[CH3NH2]
Given that Kb for CH3NH2 is a constant, you can look up its value, which is approximately 4.38×10^-4 at 25°C.
Now, we can substitute the concentrations into the equilibrium expression:
4.38×10^-4 = (x)(x) / (7.02×10^-3 - x)
Simplifying the equation:
4.38×10^-4 = x^2 / (7.02×10^-3 - x)
Next, we can solve this quadratic equation for x. Rearrange the equation to isolate x^2:
x^2 = 4.38×10^-4 (7.02×10^-3 - x)
Expand the equation:
x^2 = (4.38×10^-4)(7.02×10^-3) - (4.38×10^-4)x
x^2 + (4.38×10^-4)x - (4.38×10^-4)(7.02×10^-3) = 0
Now, we can solve this quadratic equation using either factoring, completing the square, or the quadratic formula. In this case, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
Plugging in the values, we have:
x = [-(4.38×10^-4) ± sqrt((4.38×10^-4)^2 - 4(1)(-(4.38×10^-4)(7.02×10^-3)))] / (2(1))
Solving this equation gives us two possible values for x. We need to choose the positive value since the concentration cannot be negative:
x = 7.92×10^-4
Therefore, the concentration of [CH3NH3+] in the 7.02×10^-3 M CH3NH2 solution is approximately 7.92×10^-4 M when rounded to 3 significant digits.