Determination of trace oxygen in gases
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The national standard GB/T5831-2011 for the determination of trace oxygen in gases states that 1 mL of a 0.05 mol/L copper sulfate solution is equivalent to 0.300 mL of oxygen at 20°C and 103 kPa: 0.05/1000*11.2*1000/2*293/273 = 0.300. Where does the value 11.2 come from? Isn’t it calculated using the formula n*22.4=V, that is, 0.05*2/1000*22.4*1000*293/273?
This post was last edited by chicanpk on 2016-8-16 at 14:54. The original poster can take a look at the equation for this reaction as specified in the standards: it involves 2 atoms of Cu reacting with 1/2 atom of O2. The amount of copper ions in 1 ml of copper sulfate is 0.05/1000 moles; accordingly, the amount of 1/2 O2 is 0.05/1000/2 moles. Thus, the amount of O2 is 0.05/1000/2/2 moles. The volume of O2 under standard conditions is then 0.05/1000/2/2 * 22.4 * 1000 ml, which simplifies to 0.05/1000/2 * 11.2 * 1000 ml. When converted to the volume at 20 degrees Celsius, it becomes 0.05/1000/2 * 11.2 * 1000 * 293/273 ml, or 0.300 ml. n*22.4=V is correct; the key is to calculate the value of n accurately
Please, the original poster, share this standard with us
This post was last edited by yao*aoling on 2016-8-23 at 17:09. It can be downloaded from Standard Sharing Network; thank you