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 = ?
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%
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