Archive for March, 2008
Chemistry Tutor Site Question
The balance reaction of acetic acid that reacts with sodium hydroxide is,
HC2H3O2 (aq) + NaOH (aq) ——–> H2O (l) + NaC2H3O2 (aq)
If 2.50 mL of vinegar needs 34.9 ml of 0.0960 M NaOH to reach the equivalence point in titration, how many grams of acetic acid are in a 1.00-qt sample of this vinegar?
Chemistry Tutor Site Answer
This is an example of titration weak acid with strong base. Acetic acid is a weak acid and sodium hydroxide is a strong base. At the end point the mol equivalent of acid will be equal to the mol equivalent of base. Use this formula when you solve problem that involve titration calculation:
N x V acid = N x V base
n x M x V (acid) = n x M x V ( base)
Where n is the total amount of ion H+ in acid or OH- in base, N is Normality, M is Molarity, and V is Volume. Thus for the question above we have:
1 x M x 2.50 mL = 1 x 0.0960 M x 34.9 mL
M = 1.340 M
This means that in 1 L of vinegar contains 1.340 mol of acetic acid. Multiply with its molar mass thus we can get the mass of acetic acid in the 1 L of vinegar.
mass of acetic acids = mol x molar mass
mass of acetic acids = 1.340 mol x 60 =
mass of acetic acids = 80.4 g
because 1 quart, liquid (U.S.) is equal to 0.946 liter, thus the mass of acetic acid per 1.00 qt ia
= 80.4 g /L x 0.946 L/qt
= 76.06 g/qt
Thus there are 76.06 g of acetic acid per 1 qt of vinegar
March 31st, 2008
Chemistry Tutor Site
How much of total volume in the conical flask at the end point of titration, when 0.1 M of sulphuric acid (H2SO4) is used to titrate 20 mL of 0.2 M of sodium hydroxide (NaOH)?
Chemistry Tutor Site
This is an acid-base titration that involve sulphuric acid (H2SO4) and sodium hydroxide (NaOH). H2SO4 is a strong acid and NaOH is a strong base, in the end point the pH of the solution will be 7 or neutral. The balance reaction is:
H2SO4(aq)+2NaOH(aq) –> 2H2O (l) +Na2SO4(aq)
Always use this formula when you calculate titration problem, in the end point the mole equivalent titrant is equal to the mol equivalent of titer, thus we can write:
N1V1=N2V2
n1.M1.V1 (acid) = n2.M2.V2 (base)
Where are n1 or n2 are the amount of H+ in acid or OH- in the base. In this case n1 is 2 because H2SO4 has 2 ion H+, and n2 is 1 because NaOH has one OH-, thus
n1.M1.V1 (acid) = n2.M2.V2 (base)
2×0.1xV1 = 1 x 0.2 x 20
V1 = 20 mL
So the total volume in the end point is 20 mL + 20 mL = 40 mL
March 31st, 2008
Chemistry Tutor Site - Question
In Mohr titration using Ag ion, why does AgCl and not Ag2CrO4 that precipitates first, although the Ksp value for AgCl (10-10) is greater than Ksp Ag2CrO4 ( 10-12)? If Ag2CrO4 is less soluble than AgCl why this happen when titrated with silver and chromate as indicator?
Chemistry Tutor Site – Answer
You can not use the Ksp value for determine which ones of the species are less soluble than the other, and conclude that the smaller Ksp value is will precipitate fisrt. The best way to compare the solubility of the species is using the “solubility” rather than Ksp value.
For the practical approach when you do the Mohr titration, you must see that AgCl will precipitate first rather than Ag2CrO4, this circumstance giving us a hypothesis that silver chromate is more soluble than silver chloride, but how we can make a theoretical calculation to show that our hypothesis is right? You can do this by calculating the solubility (s) for each species:
Ksp Ag2CrO4 = [Ag+]2[CrO4-2] = 1.1×10-12
let x be the molarity of CrO4-2, we know then that the molarity of the silver would be twice that value. Therefore
[2x]2[x] = 1.1×10-12 , solve for x
4x2.x = 1.1×10-12
4x3=1.1×10-12
x= 6.51×10-5 M
and therefore is the same as the number of moles of Ag2CrO4 that is in solution at the saturation point. This equal to:
=6.51×10-5 mole Ag2CrO4/liter * 331.73 g/mole
= 2.15×10-2 g, that is about 20 mg in a liter.
What about AgCL solubilty?
Ksp AgCl = 10-10, that works out to be 1.26×10-5 Molar, and the mass per litter is
= (1.26×10-5 moles/liter * 143.32g/mole )
= 1.81×10-3 g, that is almost 2 mg in a liter.
from the calculation above we approve that AgCrO4 is less soluble than AgCl and because of these AgCl will be precipitate first when we perform Mohr titration.
March 30th, 2008