6D. 19

[505316964]

A sample of CH3NH3Cl of mass 15.5 g is dissolved in water to make 450. mL of solution. What is the pH of the solution?

I keep doing this and getting a PH of 2.6, the answers say it's supposed to be 5.42.

I found the Ka to be 2.3 X 10^-5 = x^2/0.265-x

I get x to be .0025

can someone tell me what i'm doing wrong?

[Osvaldo SanchezF -1H]

The reason that you don't get the right answer is because you have your Ka as the wrong value. If you look at the table from 6C.2 you notice that the Kb of CH3NH2 is equal 3.6*10^-4 but this is a base and we are given its acidic form so all we have to do is find the Ka of the acidic form of CH3NH2 which is CH3NH3 and do that all we have to do is do Ka*Kb=Kw where we know both the Kb and Kw so we plug it in and get Ka=2.8*10^-11. Now with our new Ka you can plug it in the formula and btw you can use the 5% rule to make it easier. You should get 5.42 as the pH.