Find the pH of a 0.050 M solution of trimethylamine ((CH_3)_3N). Write out the k
ID: 495403 • Letter: F
Question
Find the pH of a 0.050 M solution of trimethylamine ((CH_3)_3N). Write out the key equilibrium first. Find K_a using the K_a given in problem 3. (Trimethylamine is the conjugate base of trimethylammonium chloride. K_w.= 1.0 times 10 M) Solve using the K_a. Start here If you mixed 100 ml of the solution In problem 1 with 100 ml of the solution in problem 2, how would it affect each equilibrium? What would the pH be? A buffer solution contains 0.100 mole of propionic acid (CH_3CH_2COOH, pK_a = 4.86) and 0.100 mole of sodium propionate (Na^+ CH_3CH_2COO) dissolved in a volume of 0.500 L. a. What is the pH? b. The amount of each component is increased by 0.010 mole. What happens to the new pH? c. The volume of the solution is doubled to 1.000 L. What happens to the pH? What is the pH of pH of a solution containing 0.25 M NH_4 and 0.40 M NH_4CI K_a of NH_4^+ = 5.70 times 10^-10 What is the pH of a solution containing 0.75 M lactic acid and 0.25 M sodium lactate? K_a = 1.4 times 10^-4 7.0.100 L of 0.100 M HCOOH (K_a = 1.8 times 10^-4.) is mixed with 0.075 of 0.075 M NaOH and diluted to 1.00 L. What is the pH of the solution?Explanation / Answer
Q2.
find pH of solution of trimethylamine (B)
a) equilibirum
(CH3)3N(aq) + H2O(l) <--> (CH3)3NH+(aq) + OH-(aq)
Kb = [ (CH3)3NH+][OH-]/[(CH3)3N]
b)
find Kb using Ka
Ka give = 1.56*10^-10
at 25°C:
Kw = Kb*Kb = 10^-14 always valid
so
(1.56*10^-10)(Ka) = (10^-14)
Kb = (10^-14)/((1.56*10^-10)) = 6.410*10^-5
c)
pH for:
Kb = [(CH3)3NH+][OH-]/[(CH3)3N]
assume, in equilibrium:
[(CH3)3NH+] = x= [OH-]; due to sotichiomerty
[(CH3)3N] = M-x = 0.05 - x (account for equilibrium)
substitute
6.410*10^-5 = x*x/(0.05-x)
x = [OH-] = 0.001758
pOH= -log([OH-]) = -log(0.001758) = 2.7549
pH = 14-2.7549
pH = 11.2451
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