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NaOH and Na_(2)CO_(3) are dissolved in 200ml aqueous solution. Now methyl orange is added in the same solution titrated and requires 2.5ml of the same HCl. Calculate the normality of NaOH & NaCO_(3). |
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Answer» `(0.5)/(200),(1.5)/(200)` With phenolphthalein, `eq_(HCl)` = `eq_(NaOH)+(1)/(2)eqNa_(2)CO_(3)` `17.5xx0.1=axx1+bxx1` With methyl organe, `eq_(HCl)=(1)/(2)eqNa_(2)CO_(3)` `2.5xx0.1=bxx1` `implies a=1.5,b=0.25` implies `[NaOH]=(1.5)/(200)M=(1.5)/(200)N` `[Na_(2)CO_(3)]=(0.25)/(200)M=(0.5)/(200)N` |
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