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Experiment Volumetric analysis - Acid base

by TheDanny
Tags: acid, analysis, base, experiment, volumetric
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TheDanny
#1
Aug14-07, 03:39 AM
P: 8
1. The problem statement, all variables and given/known data
Do not understand the question.


2. My Lab Results

Topic : Volumetric analysis - Acid base
purpose : To determine the exact concentration of a mineral acid, HXO4, and to determine the relative atomic mass of the element X.
Materials : KA1 is a mineral acid, HXO4
KA2 is a solution containing 1.70g of OH- ions per dm3.
Phenolphthalein as indicator.

Procedure : Pipette 25.0cm3 of KA2 into the titration flask. Add two or three drops pf phenolphthalein indicator and titrate this solution with KA1. Record your readings in the table.
Repeat the titration as many times as you think necessary to achieve accurate results.


My result up.

My problems is here the question.
a. Calculate the concentration, in mol dm-3, of solution KA2
b. Write a balanced ionic equation for the reaction between solution KA1 and the solution KA2.
c. Calculate the concentration, in mol dm-3, of mineral acid HX04 in solution KA1.
d. If the concentration of mineral acid HXO4 in solution KA2 is 20.1g dm-3, calculate the relative molecular mass of HXO4.
e. Using the answer to (d), determine the relative atomic mass of the element X.
f. Suggest and identity for element X.

Please guide me. Thx

In question (a) i tried with
mol=mass/mm
=1.7/17 = 0.1mol

mol=mv/1000
0.1=m(250)/1000
m=0.4
concentration of KA2 is 0.4moldm-3?
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chemisttree
#2
Aug14-07, 09:51 AM
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Quote Quote by TheDanny View Post

In question (a) i tried with
mol=mass/mm
=1.7/17 = 0.1mol
This is correct. Remember that the concentration was given as 1.70 g OH- per dm3 which is the same thing as 1.7 g OH- per liter or 0.1 moles/liter OH-, so this...
mol=mv/1000
0.1=m(250)/1000
m=0.4
concentration of KA2 is 0.4moldm-3?
... calculation isn't necessary.

For the rest of the problem, begin by understanding the neutralization equation.
Try to write the neutralization equation using [tex]M^+OH^-[/tex](monobasic) and [tex]HXO_4[/tex](monoprotic).


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