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200 " mL of " a solution of a mixture of...

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.

Text Solution

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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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Knowledge Check

  • A mixture NaOH+Na_(2)CO_(3) required 25mL of 0.1 M HCl using phenolpththalein as the indicator. However, the same amount of the mixture required 30mL of 0.1M HCl when methyl orange was used as the indicator. The molar ration of NaOH and Na_(2)CO_(3) in the mixture was:

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    `2:1`
    B
    `1:2`
    C
    `4:1`
    D
    `1:4`

  • A 0.1 N solution of Na_(2)CO_(3) is titrated with 0.1 N HCl solution. The best indicator to be used is

    A
    Potassium ferricyanide
    B
    Phenolphthalein
    C
    Methyl red
    D
    Litmus Paper

  • 3.65gm equimole mixture of NaOH and Na_(2)CO_(3) is titrated against 0.1M HCl using phenolphathalein as an indicator, V_(1)mL of acid was required to reach end point. In another experiment 3.65gm of same mixture is titrated against 0.2M HCl using methyl orange as an indicator, V_(2)mL of acid was required to reach end point. V_(1)+V_(2) is :

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    B
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    C
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    D
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