Suppose you have a 100.00ml of 0.01000M methylamine (CH3NH2) solution ( K b = 4.
ID: 884655 • Letter: S
Question
Suppose you have a 100.00ml of 0.01000M methylamine (CH3NH2) solution (Kb = 4.4×10-4).
(a) Calculate the initial equilibrium pH of this solution.
(b) Calculate the equilibrium pH after the addition of 70.00mL of 0.01000M HCl to methylamine using both the chemical equilibrium approach and the Henderson-Hasselbalch equation approach? Explain why there is a difference between the two calculated values.
I found the ph through the HHB method but i can't get the right answer for the equilibrium method. The answer I am supposed to get for part B is 10.21.
Here is what I did to solve for part B.
CH3NH2 + HCl = Ch3NH3 + Cl-
S .001 0.0007 0
R -.0007 -0.0007 0.0007
F 0.0003/(.17) 0 0.0007
C .001764706 0 .004117647
pH=-log(2.27*10^-11)+[ .001764706/ .004117647]
pH=10.28
Explanation / Answer
CH3NH2 + H2O ---------------------> CH3NH3+ + OH-
0.01 0 0 -----------------> initial
0.01-x x x ---------------> equilibrium
Kb = [CH3NH3+][OH-]/[CH3NH2]
4.4 x 10^-4 = x^2 / 0.01-x
x^2 + 4.4 x 10^-4 x - 4.4 x 10^-6 = 0
x = 1.89 x 10^-3
[OH-] = x = 1.89 x 10^-3 M
pOH = -log[OH-]
pOH = -log (1.89 x 10^-3 )
pOH = 2.72
pH+ pOH = 14
pH = 11.28
(b)
millimoles of HCl = 0.01 x 70 = 0.7
millimoles of CH3NH2 = 0.01 x 100 = 1
CH3NH2 + HCl ----------------> CH3NH3+Cl-
1 0.7 0 ---------------------> initial
0.3 0 0.7 --------------------> equilibrium
base and salt reamins in the mixture . it can form buffer
pKb = -log (4.4 x 10^-4) = 3.36
pOH = pKb + log [salt/base]
pOH = 3.36 + log [0.7/0.3]
pOH = 3.73
pH = 10.27
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