help, i'd like to know how to do this titration question:
for example: a 1g sample of naoh contaminated with na2co3 is titrated with 0.500MHCl, and it is found that 46.5mL of HCl are required for neutralization. Find the percentage by mass Na2Co3.
thank you in advance
Did you make this up from scratch? The numbers don't look right to me.
no, this is from my lab book. and furthermore, i am not able to solve it.
To find the percentage by mass of Na2CO3 in the NaOH sample, we need to use the concept of titration and stoichiometry.
First, let's set up the balanced chemical equation for the reaction between NaOH and HCl:
2NaOH + HCl -> Na2CO3 + 2H2O
From the equation, we can see that the mole ratio between NaOH and Na2CO3 is 2:1.
Now, let's calculate the number of moles of HCl that reacted in the titration:
Volume of HCl = 46.5 mL = 0.0465 L (since 1 mL = 0.001 L)
Molarity of HCl = 0.500 M
Number of moles of HCl = Molarity x Volume
= 0.500 M x 0.0465 L
= 0.02325 moles
Since the mole ratio between NaOH and HCl is 1:1, the number of moles of NaOH is also 0.02325 moles.
Now, using the mole ratio between NaOH and Na2CO3, we can find the number of moles of Na2CO3:
Number of moles of Na2CO3 = (1/2) x 0.02325 moles
= 0.01163 moles
Next, let's calculate the molar mass of Na2CO3:
Molar mass of Na2CO3 = (2 x atomic mass of Na) + atomic mass of C + (3 x atomic mass of O)
= (2 x 22.99 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol)
= 105.99 g/mol
Now, let's calculate the mass of Na2CO3 in the sample:
Mass of Na2CO3 = Number of moles x Molar mass
= 0.01163 moles x 105.99 g/mol
= 1.23 g
Finally, let's calculate the percentage by mass of Na2CO3 in the sample:
Percentage by mass of Na2CO3 = (Mass of Na2CO3 / Mass of sample) x 100%
= (1.23 g / 1 g) x 100%
= 123%
Therefore, the percentage by mass of Na2CO3 in the NaOH sample is 123%.