Question #5b2c0
1 Answer
Explanation:
This is a pretty straightforward practice problem on using an ICE table to figure out the equilibrium concentration of the hydroxide ions.
Calculate the initial molarity of methylamine by using the number of moles and the volume of the solution
#color(blue)(c = n/V)#
#c = "0.10 moles"/"1.0 L" = "0.10 M"#
Before moving on to the actual ICE table, take a look at the value of the base dissociation constant,
When
So, an ICE table for this reaction would look like this
#"CH"_3"NH"_text(2(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "CH"_3"NH"_text(3(aq])^(+) " "+" " "OH"_text(aq])^(-)#
By definition, the base dissociation constant will be equal to
#K_b = ( ["OH"^(-)] * ["CH"_3"NH"_3^(+)])/(["CH"_3"NH"_3])#
In your case, this will be equal to
#K_b = (x * x)/(0.10 -x ) = x^2/(0.10 - x)#
Because
#0.10 -x ~~ 0.10#
This means that you have
#K_b = x^2/0.10 = 4.4 * 10^(-4)#
This will get you
#x = sqrt(0.10 * 4.4 * 10^(-4)) = 6.6 * 10^(-3)#
Since