Posts filed under 'Titration'
Chemistry Help - Chemistry Problem
A solid White sample of Zn(OH)2 is added to 0.300 L of 0.420 M HBr. solution. The solution that remains is still acidic and It is then titrated with 0.420 M NaOH solution, and it takes 94.5 mL of the NaOH solution to reach the equivalence point.
What mass of Zn(OH)2 was added to the HBr solution?
Chemistry Tutor Answer
To solve this problem you only need to find how many moles of HBr that reacted with Zinc hydroxide Zn(OH)2 How you will find it? First you must find the initial moles of HBr and than subtracts with the moles of HBr that reacted with natrium hydroxide.
After that change the mol of HBr into moles of Zn(OH)2 using the coefficient of the reaction and multiply with the molecular mass to obtain the mass of zinc hydroxide. here are the problem solving of that chemistry problem.
moles of HBr initial
= 0.300 L x 0.420
= 0.126 mol
mol of NaOH
= 0.420 M x 94.5 mL
= 39.69 mmol
= 0.039 mol
NaOH reacted with HBr according this reaction, remember to find the mol of HBr you only need to compare the reaction coefficient since the coefficient of HBr and NaOH are equal thus their moles would be the same.
NaOH + HBr -> NaBr + H2O
from the reaction the mol of HBr are
= 1/1 x 0.039 mol
= 0.039 mol
thus the mol of HBr that reacted with Zn(OH)2 are
= Mol HBr initial - mol HBr that reacted with NaOH
= 0.126 mol - 0.039 mol
= 0.087 mol
Zn(OH)2 reacted with HBr according this reaction to give ZnBr2 and H2O, remember to find the mol of Zn(OH)2 you only need to compare the reaction coefficient since the coefficient of HBr and Zn(OH)2 are 2:1 thus the moles of Zn(OH)2.
Zn(OH)2 + 2 HBr -> ZnBr2 + 2 H2O
from the reaction we can find moles of Zn(OH)2
= 1/2 x moles of HBr
= 1/2 x 0.087 mol
= 0.0435 mol
and the mass
= mol x mr
= 0.0435 mol x 99.396
= 4.324 g
October 9th, 2008
Standardization denotes a process to determine the concentration of a solution using a standard solution. Base usually has a hydroscopic property because of this we can not use base like this for a primer standard. We use another substance to standardize this solution. using this technique we have a base standard solution that can be reacted to determine acid solution. take a look for the example
A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.01 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 36.5 mL of potassium hydroxide to reach the endpoint. What is the molarity of the potassium hydroxide solution? Same solution used to titrate an unknown solution of hydrochloric acid, if 22.1 mL of the potassium hydroxide solution is required to neutralize 21.9 mL of HCl, find molarity of HCl?
First calculate the moles of potassium hydrogen phthalate KHC8H4O4
= mass/ molar mass
= 1.01 g / 204.2
= 4.946 x 10-3 mol
= 4.946 x 10-3 equivalents
= 4.946 milli-equivalent
to calculate the normality of KOH we used titration formula
V x N acid = V x N base
eq acid = V x N base
4.946 = 36.5 mL x N
N = 0.136 N
Potassium hydroxide has one ion hydroxide thus 0.136 N of KOH should be 0.136 M of KOH. By the same way we can calculate the molarity of HCl,
V x N acid = V x N base
21.9 x N = 22.1 x 0.136 N
N HCl = 0.137 N
As KOH, HCl has one hydrogen ion thus 0.137 N of HCl = 0.137 M of HCl.
June 20th, 2008
Back titration can be use to determine the concentration of unknown acid or base. The unknown concentration of the base solution firstly reacted with the excess acid and then the excess can be titrated with the standard base solution, from the difference of the moles of the acid we can determine the concentration of the unknown base. Here is the example:
38.7 mL of 0.514 M nitric acid is added to 20.7 mL of calcium hydroxide, and the resulting solution is found to be acidic. 27.0 mL of 0.411 M potassium hydroxide is required to reach neutrality. What is the molarity of the original calcium hydroxide solution?
Answer:
moles of HNO3 original = 38.7 mL x 0.514 = 19.892 mmol
mmol of HNO3 that reacted with KOH is calculated using titration formula, because KOH has one hydroxide ion thus 0.411 M KOH would be equal to 0.4111 N of KOH
meq acid = meq base
meq acid = V x N
meq acid = 27.0 mL x 0.411 N
meq acid = 11.097 milli-equivalent
because HNO3 has one ion hydrogen thus 11.097 milli-equivalent of HNO3 are equal to 11.0971 mmol HNO3. The moles of HNO3 that reacted with Ca(OH)2 can be calculated by subtracting the origin moles of HNO3 with the moles of HNO3 that reacted with KOH, and these moles are represent the moles of Ca(OH)2 too.
moles HNO3 that reacted with Ca(OH)2 are
= moles of HNO3 original - moles HNO3 reacted with KOH
= 19.892 mmol - 11.0971 mmol
= 8.795 mmol
Molarity of Ca(OH)2 solution are
M = mmol / mL
M = 8.795 mmol / 20.7 mL
M = 0.425 M
The molarity oh calcium hydroxide Ca(OH)2 that calculated using back titration from above calculation are 0.425 M.
June 20th, 2008
We can calculate the mass percentage of a substance using titration method. Titration formula is used to determine the mole-equivalent of the substance, if the mole equivalent of the substance is known we convert it into moles and then multiplying with molar mass to get the mass of the species. Dividing the actual mass of the substance with its original weight we have its mass percentage. Here the example
A 12.1 g sample of an aqueous solution of perchloric acid contains an unknown amount of the acid. If 20.4 mL of 0.450 M barium hydroxide is required to neutralize the perchloric acid, what is the percent by weight of perchloric acid in the mixture?
First we calculate the mole of the perchloric acid (HClO4) using titration formula. Remember V x N (volume multiply with normality produce mole-equivalent or meq), and we convert molarity of Ba(OH)2 solution into normality because Ba(OH)2 has two hydroxide ions thus 0.450 M of barium hydroxide is equal to 0.9 N.
V x N acid = V x N base
meqacid = V x N base
meq acid = 20.4 mL x 0.9 N
meq scid = 18.36 milli-equivalent
because HClO4 has one ion hydrogen thus one equivlent of HClO4 is equal to one mole of HClO4, then 18.36 milli-equivalents of HClO4 = 18.36 millimoles of HClO4. The mass of HClO4 can be calculate as
mass of acid (HClO4)
= moles x molar mass
= 18.36 mmol x 100.458
= 1844.409 mgr
= 1.844409
then the mass percentage of HClO4 are
= mass of HClO4 / mass of sampel x 100%
= 1.844409/12.1x 100%
= 15.243 %
from these calculation we get that the mass percentage of perchloric acid in the sample using titration calculation are 15.243%
June 20th, 2008
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