Calculate the weight of borax required to give a titre of 20.00mL with 0.1M HCl. Borax is sodium tetraborate decahydrate (Na2B4O7.10H2O) and we will use the analytical reagent (AR) grade, which we can assume is 99.9% pure.
All working out would be much appreciated!
All working out would be much appreciated!
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Since you must dissolve the borax to do the titration, the only way to know the weight of the borax is to know the exact concentration. Therefore the 100 mL must be measured accurately.
The reaction is
2HCl + Na2B4O7 ----> H2B4O7 + NaCl
2 moles of HCl react with 1 mole of tetraborate. 20.00 mL of 0.1 M HCl contains 0.002 mol HCl, so the amount of tetraborate reacted is 0.001 mol. The molar mass of Na2B4O7⋅10H2O is 381.4 g/mol
You need 0.3814 g
The reaction is
2HCl + Na2B4O7 ----> H2B4O7 + NaCl
2 moles of HCl react with 1 mole of tetraborate. 20.00 mL of 0.1 M HCl contains 0.002 mol HCl, so the amount of tetraborate reacted is 0.001 mol. The molar mass of Na2B4O7⋅10H2O is 381.4 g/mol
You need 0.3814 g