Write the balanced chemical equation for the precipitation of silver chlor.de fr
ID: 1047884 • Letter: W
Question
Write the balanced chemical equation for the precipitation of silver chlor.de from silver ion and chlor.de ion. What errors in experimental procedure could cause systematic errors? List one error that leads to a higher than expected result, and list one error that leads to a lower t an expect result. A solution of 0.2 M AgNO_3 will be provided. If you wish to obtain approximately 0.30 g.of AgCl from the precipitation of soluble chlorides with AgNO_3. approximately what volume of 0.2 M AgNO_3 is needed. A solution of 1.0 M HNO_3, will be provided. What other solution(s) will you need to prepare in order to complete this lab? How much will you need and how will you prepare it (them)?Explanation / Answer
1) The balanced chemical equation for the precipitation of silver chloride is
Ag+ (aq) + Cl- (aq) ---------> AgCl (s)
2) Systematic errors are errors which can be measured and infact, can be avoided. A few possible systematic errors in the gravimetric estimation of chloride as silver chloride are:
(i) loss in weight of precipitates while transferring.
(ii) overwashing or underwashing of precipitates.
(iii) incomplete drying of precipitates.
(iv) use of uncalibrated glassware in the experiment.
A possible systematic error that can lead to a higher than expected result is the incomplete drying of the precipitate. Some amount of water will be trapped in the precipitate and the moist precipitate will register and higher weight than expected.
An error that can lower the result below the expected value will be the loss of precipitate during transferring from one glassware to another. If a small amount of the precipitate is lost and there is no way to retrieve the precipitate, then we shall get a lower weight of the precipitate than expected.
3) First write down the precipitation reaction with an alkali metal chloride (say). Alkali metal chlorides are soluble in water and hence we can safely take NaCl as an example.
AgNO3 (aq) + NaCl (aq) -------> AgCl (s) + NaNO3 (aq)
We see that there is a 1:1 molar ration between AgNO3 and AgCl.
We obtain 0.30 g AgCl in the reaction. Molar mass of AgCl = 143.32 g/mol; therefore, mole(s) of AgCl precipitated = (0.30 g AgCl)*(1 mole AgCl/143.32 g AgCl) = 2.0932*10-3 mole.
As there is a 1:1 molar ratio between AgCl and AgNO3, hence the mole(s) of AgNO3 required = (2.0932*10-3 mole AgCl)*(1 mole AgNO3/1 mole AgCl) = 2.0932*10-3.
We are provided with a 0.2 M AgNO3 solution; therefore, the volume of AgNO3 required = (2.0932*10-3 mole)/(0.2 mole/L) = 0.01047 L = (0.01047 L)*(1000 mL/1 L) = 10.47 mL 10.5 mL (ans).
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