To find the pH of 0.050 M diethylamine we must use the Kb of diethylamine.The Kb value of diethylamine is as follows:Kb=1.3X10^-3Kb = [(C2H5)2NH2+][OH-]/[(C2H5)2NH] as diethylamine dissolves, some will react with water forming diethylammonium ion and hydroxide, which is not accounted for in the Kb equation. Because of this, we don't know the concentration of each ion, but we do know they will be the same. We'll call them x. The diethylamine concentration will then be 0.050 - x.Kb = x^2/(0.050 - x)1.3x10^-3 = x^2/(0.050 - x)since 1.3 x 10^-3 is not much smaller compared to 0.050, x will be significant so we cannot ignore it. Rearrange the equation and use the quadratic formula to solve for x..000065 - .0013x = x^20 = x^2 + 0.013x - .000065x = [OH]pOH = -log([OH])14 - pOH = pH
12
Yes because 0050 is the same as 50
YEs
diethylamine
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
-log(0.1 M) = 1 pH
acid base
- log(0.001 M NH4Cl) = 3 pH =====
pH=1
-log(0.5 M HF) = 0.3 pH
- log(0.12 M) = 0.92 pH ---------------
- log(0.000626 M H2SO4) = 3.2 pH -----------
- log(0.00450 M HCl)= 2.3 pH=======