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200 " mL of " a solution of a mixture of NaOH and `Na_2CO_3` was first titrated with 0.1 M HCl using phenolphthalein indicator. 17.5 " mL of " HCl was required for the same HCl was again required for next end point. Find the amount of NaOH and `Na_2CO_3` in the mixture. |
Answer» With phenolphthalein `m" Eq of "Hcl` used for 200 mL solution`=17.5xx0.1xx1` `=1.75` Let a and b m" Eq of "NaOH and `Na_(2)CO_(3)` respectively. `a+(b)/(2)=1.75` .(i) With methyl orange indicator m" Eq of "HCl used for solution after first end point `=2.5xx0.1xx1=0.25` `(b)/(2)=0.25` .(ii) By equations (i) and (ii) `a(mEq. of NaOH)=1.5` Weight of NaOH`=1.5xx10^(-3)xx40` `=(0.06g)/(200mL)` `b(m" Eq of "Na_(2)CO_(3))=0.5` Weight of `Na_(2)CO_(3)=0.5xx10^(-3)xx(106)/(2)` `=(0.0265)/(200mL)` |
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