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40 ml of 0.05 M solution of sodium sesquicarbonate (Na2 CO3 .NaHCO3 .2H2 O) is titrated against 0.05 M HCl. x ml. of HCl is used when phenolphthalein is indicator and y ml. of HCl is used when methyl orange is the indicator in two separate titrations. Hence (y – x) is
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so you have 40 mL of 0.05M NaHCO3 using molarity=moles/V, you find that there is 0.002 moles are you trying to find how many mL of HCl you used?
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