Posts filed under 'Titration'
Free chemistry Tutor Homework Help
Consider the titration of 20.0 mL of 0.0800 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HClO4.? Calculate the pH after the following volumes of titrant have been added:
(a) 0.0 mL
(b) 4.0 mL
(c) 8.0 mL
Answer
(a). Since no titrant has been added to the analyt so we only have 20.0 mL of 0.0800 M (CH3)3N thus the pH of this solution is calculated using this equation:
[OH-] = (Kb x M)1/2
[OH-] = (6.4.10-5x 0.0800)1/2
[OH-] = 2.26×10-2
pOH = - log 2.26×10-2 = 2.64
pH = 14- 2.64 = 11.36
b. 20.0 mL of 0.0800 M (CH3)3N is equal to 1.6 mmol of (CH3)3N and we add 4.0 mL HClO4 0.100 M that equal to 0.4 mmol. Thus 1.6 mmol (CH3)3N to be react with 0.4 mmol HClO4 this will be produce 0.4 mmol of (CH3)NHCLO4. The solution contains base and its salt, so this will be form a buffer solution
The balance reaction after with mole after reaction occur:
(CH3)3N + HClO4 -> (CH3)3NHClO4
1,2 mmol—0 mmol———0.4 mmol
We can use the henderson-Hasselbalch equation to calculate pH of the buffer solution
pH = pKa + log (base/salt)
Ka for (CH3)3N = Kw/Kb = 1×10-14/6.4.10-5 = 1.56.10-10
pKa = - log Ka = - log(1.56.10-10) = 9.8
pH = 9.8 + log (1.2 mmol/0.4 mmol) = 10
c. Using the same way that we use to solve the b problem we get:
(CH3)3N + HClO4 -> (CH3)3NHClO4
0.8 mmol—-0 mmol—-0.8 mmol
pH = 9.8 + log (0.8 mmol/0.8 mmol) = 9.8
November 5th, 2008
Free Chemistry Tutor Homework help
How many milliliters of 0.861 M HBrO4 are needed to titrate each of the following solutions to equivalence point?
- 75.0 mL of 1.46 M LiOH
- 48.1 mL of 1.55 M NaOH
- 281.0 mL of a solution that contains 44.2 g of RbOH per liter
Answer
For titration problem, there is a formula that you can use to solve titration problem. This formula is uneversal for all kind of titration, and the formula is:
V x N (acid) = V x N (base)
V is the volume and N is the normality. For the problem above we need to calculate the normality of all the solution. Just for remember, to change from molarity into normality for acid or base you just multiply the molarity with the number of proton (acid) or the number of OH- for base.
0.861 M HBrO4 = 0.861 N HBrO4
1.46 M LiOH = 1.46 N LiOH
1.55 M NaOH = 1.55 N NaOH
Normality of 281.0 mL of a solution that contains 44.2 g of RbOH are:
mole = 44.2 / 102.48 = 0.431 mol
M = 0.431 mol/0.281 L = 1.53 M = 1.53 N
1. 75.0 mL of 1.46 N LiOH
75.0 x 1.46 N = 0.861 N x V
V = 127,2 mL
2. 48.1 mL of 1.55 N NaOH
48.1 x 1.55 N = 0.861 N x V
V = 86.6 mL
3. 281.0 mL of 1.53 N RbOH
281.0 x 1.53 N = 0.861 N x V
V = 499.3 mL
November 5th, 2008
Chemistry Tutor Site - Chemistry Homework Help
A laboratory technician wants to determine the aspirin content of a headache pill by acid-base titration. Aspirin has a Ka of 3.0 x 10-4 M. If the pill is dissolved in water to give a solution about 0.005 M, what is the pH of this solution? (Neglect dilution effects.)
If the solution in the problem above is then titrated against KOH solution, what will be the pH at the stoichiometric point.
Chemistry Tutor Site - Chemistry Homework Answer
With this Ka and concentration, you can’t use the usual formula for weak-acid equilibrium
HA -> H+ + A-
HA dissociated into H+ and A-, if the X mol of the HA is dissociated thus the H+ and A- would be X mol too. Thus
X2 / [Acid conc] = Ka
because you can’t neglect the amount of acid actually reacted to H+ and the Cation. So you have the more general equation
X2 /[0.005-X]= 3×10-4
You can solve this by trial/error or by the general quad. eqtn method. Roughly, X=0.0027
At the stoichiometric point, all acid is dissociated, and the problem is equivalent to dissolving 0.005 M potassium salicylate in water. In that case,
Salicylate- + H2O = Acid + OH-
In this situation, the equilibrium expression is
[Acid][OH-]/[Salicylate] = Kb
Since KbKa=10×10-15, Kb= 3.3×10-11/p>
If X is the amount of [OH-] formed at equil., then
X2/[0.005]= 3.3×10-11
X is roughly 4×10-7. From this you can compute pOH= 7-log 4= 6.4. Then since pH+pOH=14, pH=7.6.
October 17th, 2008
Chemistry Tutor Site - Chemistry Homework Help
Why does the pH drop rapidly when an acid-base titration reaches equivalence point? Why doesn’t the titration curve look more gradual? The acid is being added drop by drop to the base, so shouldn’t the pH fall drop gradually? Why does the PH change so sharply when it reaches the equation ratios? What is going on at an aqueous level?
Chemistry Tutor Site - Chemistry Homework Answer
The weaker the acid or base is the more gradual the titration curve will look. You can see this if you compare the titration curve of acetic acid to that for hydrochloric acid. Acetic acid has a much more gradual curve.
When you are at the equivalence point the pH is about 7 (depending on the acid/base), which means the hydronium ion concentration is 10^-7. If you add an acid or base it will move away from that concentration by orders of magnitude, which causes the sharp up or down (depending what is being tritrated). After a few drops have been added in the titration the curve levels out because you are only doubling or adding to the ion concentration at the same magnitude instead of changing the order of magnitude.
In other words, after the equivalence point is reached, when you add a drop the pH would change from 10^-7 to 10^-6 (if titrating with an acid) or 10^-8 (if titrating with a base. This is a factor of 10 change, and the curve is steep. But later on if you add another few drops you only change the concentration by doubling the ion concentration. So titrating with an acid goes from, say, 1×10^-5 to 2×10^-5 or something similar if titrating with a base.
October 17th, 2008
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