Chemistry - Dilution in volumetric analysis
Dilution in volumetric analysis
Some analyses using titration involve dilution of the solution being
analysed. The need to dilute is generally determined during the design of
In order to check the validity of a manufacturer's claim that a particular
cleaning agent contains 1.7% (m/v) NH3, an appropriate analysis would be
titration with hydrochloric acid solution, say 0.1000 M.
* good titration technique is to aim for a titre of around 20 mL
* 20 mL of 0.1000 M HCl(aq) - 0.1000 x 20x10 = 0.0020 mol HCl
* Since NH3 and HCl react in 1 to 1 mole ratio: NH3(aq) + HCl(aq) -
NH(aq) + Cl(aq), there would have to be approximately 0.0020 mol NH3 in the
aliquot of cleaning agent used in the analysis.
* The cleaning agent is 1.7 % (m/v) NH3, ie it contains 1.7 g NH3 in 100
* This is equivalent to 17 g NH3 per L and, since _M_(NH3) = 17 g mol, 1
mol NH3 per L.
* Aliquot size can be determined by V =_ n / c_ = 0.0020 mol / 1 mol L =
0.0020 L = 2.0 mL However, an aliquot of 2.0 mL would increase the impact
of the errors associated with the analysis.
If the cleaning agent is diluted by a factor of 10, by making 20 mL up to
200 mL with water in a 200 mL volumetric flask, a 20 mL aliquot of the
diluted cleaning agent should contain the same_ n_(NH3) as 2 mL of the
original cleaning agent.
We still have the required (approximately) 0.0020 mol NH3 but the volumes
of cleaning agent and diluted cleaning agent measured would be 20 mL, thus
reducing the impact of any errors.
An important component of the calculations associated with this dilution
is the recognition that
_n_(NH3) in 20 mL cleaning agent
=_ n_(NH3) in 200 mL of diluted cleaning agent
= [_n_(NH3) in 20 mL diluted cleaning agent / 20] x 200
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