A solution is made by dissolving 7.50g of NaOH, which contains some impurities in water and making up the solution to 250cm^3 of the solution. If 25.0cm^3 of the solution is exactly neutralized by 14.0cm^3 of nitric acid of concentration 1.00mol/dm^3, calculate the percentage purity of the sodium hydroxide.
2007-08-11 16:24:17 · 3 answers · asked by Stormy Knight 1 in Science & Mathematics ➔ Chemistry
One mol of NaOH is needed to react with one mol of HNO3. Then, 14 cm3 of acid are equivalent to 0.0140 moles of Na OH.Now , 7.50 g of NaOH =0.1875 moles of NaOH ,if it is pure, but we know that if this was true it will be necessary 0.1875/10 moles of HNO3 to neutralize 25 cm3 of Na OH.We Know now that the original solution has only 0.014 x 10 moles of NaOH=5.60 g NaOH pure.
Then % purity=(5.60/7.50)x100=74.67
2007-08-11 17:27:35 · answer #1 · answered by vega 2 · 0⤊ 0⤋
Compute the moles of HNO3 in the nitric acid solution.
Compute 10x that number of moles and multiply by the mole weight of NaOH (the 10x is involved since the titration used 1/10 of the solution of NaOH).
Divide the weight found by 7.5 g. Roughly, I get about 60 percent purity.
2007-08-11 23:34:10 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
You are assuming impurities do not partake in reaction. This is bad science and without knowing what the impurity if it can not be calculated.
2007-08-11 23:32:44 · answer #3 · answered by Anonymous · 0⤊ 0⤋