Calculate the mass of sodium trioxocabonate(IV)

Solution A is 0.10moldm-3 HCl. B is a solution of ... Solution A is 0.10moldm-3 HCl. B is a solution of sodium trioxocarbonate(IV). 25cm3 of B is titrated against A using methyl orange as indicator. 23.5cm3 of A is used in titration. Calculate the mass of sodium trioxocabonate(IV) present in one dm3 of the solution B. [Na = 23, C = 12, O = 16]

molar concentration of acid, CA =[math]0.10moldm^{-3}[/math]

volume of acid, VA = [math]23.5cm^3=0.0235dm^3[/math]

molar concentration of base, CB=?

volume of base, VB = [math]25cm^3=0.025dm^3[/math]

from equation of reaction:

[math]HCl(aq)+Na_2CO_3(aq)\to2NaCl(aq)+H_2O(l)+CO_2(g)[/math]

we can see that

a/b = mole ratio of acid to base

[math]\therefore[/math] a:b = 2:1

[math]\to[/math]

Solution A is 0.10moldm-3 HCl. B is a solution of ... Solution A is 0.10moldm-3 HCl. B is a solution of sodium trioxocarbonate(IV). 25cm3 of B is titrated against A using methyl orange as indicator. 23.5cm3 of A is used in titration. Calculate the mass of sodium trioxocabonate(IV) present in one dm3 of the solution B. [Na = 23, C = 12, O = 16]

molar concentration of acid, CA =[math]0.10moldm^{-3}[/math]

volume of acid, VA = [math]23.5cm^3=0.0235dm^3[/math]

molar concentration of base, CB=?

volume of base, VB = [math]25cm^3=0.025dm^3[/math]

from equation of reaction:

[math]HCl(aq)+Na_2CO_3(aq)\to2NaCl(aq)+H_2O(l)+CO_2(g)[/math]

we can see that

a/b = mole ratio of acid to base

[math]\therefore[/math] a:b = 2:1

[math]\to[/math]

[math]CB=\frac{bCAVA}{aVB}=\frac{0.10\times0.0235}{2\times0.025}=0.047moldm^{-3}[/math]

molar mass of [math]Na_2CO_3=(23\times2)+12(16\times3)=106gmol^{-1}[/math]

The problem now is how do I get the mass of sodium trioxocarbonate(IV) present in one [math]dm^3[/math] of the solution B?

Moderator please help me remove the other two redundant threads

Edited August 21, 201411 yr by Chikis