2gm of mixture hydrated sodium carbonate na2c03 /OH20 and sodium bicarbonate was dissolved in water and made up to 250cc. 25cc of this solution was titrated, using methyl orange as indicator and 22.5cc of 0.087N hcl were required for neutralization. Calculate the percentage of sodium bicarbonate in the mixture? The right answer 56.7%
Who helps me step by step Please
mass sample titrated = 2 g x (25/250) = 0.2 g.
You have two unknowns which will require two equations and you solve those simultaneously.
Let X = mass Na2CO3
and Y = mass NaHCO3
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Equation 1 is :
X + Y = 0.2
Then # milliequivalents = grams/milliequivalent weight = g/mew.
mew Na2CO3 = molar mass/2000 = 106/2000= 0.053
mew NaHCO3 = molar mass/1000 = 0.084
Equation 2 is
#milliequivalents Na2CO3 + #milliequivalents NaHCO3 = #milliequivalents HCl = mL x N = 22.5 x 0.087 = about 1.957 or in terms of X and Y it is:
(X/0.053) + (Y/0.084) = 1.957
Solve for Y which is in grams, then
%Y = (mass Y/mass sample)*100 = ?
To calculate the percentage of sodium bicarbonate in the mixture, you need to determine the amount of sodium bicarbonate that reacted during the titration.
Let's break down the given information and calculate step by step:
1. Calculate the number of moles of HCl used in the titration:
- Volume of HCl used = 22.5 cc (cubic centimeters)
- Concentration of HCl = 0.087N (normality)
- Convert cc to liters: 22.5 cc / 1000 = 0.0225 L
- Moles of HCl used = concentration * volume = 0.087N * 0.0225 L = 0.0019625 moles
2. Determine the balanced chemical equation for the reaction between HCl and sodium bicarbonate (NaHCO3):
NaHCO3 + HCl -> NaCl + H2O + CO2
3. From the balanced equation, we can see that one mole of HCl reacts with one mole of NaHCO3. Therefore, the number of moles of NaHCO3 in solution is also 0.0019625 moles.
4. Calculate the molecular weight of NaHCO3:
- Molecular weight of NaHCO3 = (Na: 22.99) + (H: 1.01) + (C: 12.01) + (O: 16.00 x 3) = 84.01 g/mol
5. Calculate the mass of NaHCO3 in the sample solution:
- Mass of NaHCO3 = moles * molecular weight = 0.0019625 moles * 84.01 g/mol = 0.16496 g
6. Now, we need to relate the mass of NaHCO3 to the initial mass of the mixture. From the given information, we know that the mixture weighed 2 g in total.
7. Calculate the percentage of NaHCO3 in the mixture:
- Percentage of NaHCO3 = (mass of NaHCO3 / mass of mixture) * 100
- Percentage of NaHCO3 = (0.16496 g / 2 g) * 100 = 8.248%
Based on the calculations, the percentage of sodium bicarbonate (NaHCO3) in the mixture is approximately 8.248%. The given answer of 56.7% seems to be incorrect. Please recheck the problem or provide any additional information if available.