A student titrates a sample with HCl solution using methyl red as indicator and finds it contains 20% K2CO3. Later he realises his sample contained Na2CO3 instead of K2CO3. What is the % concentration of Na2CO3 in the sample?
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Expert's answer
2016-01-19T09:48:04-0500
According to the given conditions the sample can contain the same number of moles of K2CO3 or Na2CO3. Let’s assume that he has 100 g of the sample. Then m(K2CO3) = (20% × 100 g)/100% = 20 g At the same time the amount K2CO3 is: ν(K2CO3) = m(K2CO3)/M(K2CO3) = 20 /138 mol = 0.1449 mol When he realizes the he has got Na2CO3: ν(K2CO3) = ν(Na2CO3) =m(Na2CO3)/M(Na2CO3) = 0.1449 mol m(Na2CO3) = 0.1449 mol × 106 g/mol = 15.36 g And, finally, the concentration of Na2CO3 is: W(%) = m(Na2CO3)×100%/100 g = 15.36 %
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