Question
A solution of sodium bicarbonate is prepared by adding 45.00g of sodium bicarbonate to a 1.00-L volumetric flask and adding distilled water until it reaches the 1.00-L mark. Calculate the concentration of sodium in units of:
a) milligrams per liter
b) molarity
c) normality
d) milligrams per liter as CaCO3
a) milligrams per liter
b) molarity
c) normality
d) milligrams per liter as CaCO3
Answers
sodium bicarbonate is NaHCO3.
45.00 g/L. Convert to mg/mL. To convert g to mg, multiply by 1000. To convert from L t mL, multiply by 1000. So 45.00 g/L is
45.00 g/L x (1000 mg/g)(1 L/1000 mL) = 45.00 mg/mL for NaHCO3. Since there is 1 atom Na per molecule of NaHCO3, that is the concn of Na^+ in mg/mL.
b. Convert g NaHCO3 to moles. That will be moles Na^+ also. Then moles/L = M.
c. There is 1 H in NaHCO3, therefore, N = M.
d. Convert 45.00 g NaHCO3/L to g CaCO3/L, then change to mg/L.
Post your work if you get stuck.
45.00 g/L. Convert to mg/mL. To convert g to mg, multiply by 1000. To convert from L t mL, multiply by 1000. So 45.00 g/L is
45.00 g/L x (1000 mg/g)(1 L/1000 mL) = 45.00 mg/mL for NaHCO3. Since there is 1 atom Na per molecule of NaHCO3, that is the concn of Na^+ in mg/mL.
b. Convert g NaHCO3 to moles. That will be moles Na^+ also. Then moles/L = M.
c. There is 1 H in NaHCO3, therefore, N = M.
d. Convert 45.00 g NaHCO3/L to g CaCO3/L, then change to mg/L.
Post your work if you get stuck.
a) 45.00g/L(1000mg/g)(1L/1000mL)=45mg/mL
b)NaHCO3: 22.99+1+12+3(16)= 83.99g/mol
prop. of Na (22.99/83.99)=0.54
c)?
d)?
b)NaHCO3: 22.99+1+12+3(16)= 83.99g/mol
prop. of Na (22.99/83.99)=0.54
c)?
d)?
a is ok.
b you have close to the correct answer but not the correct procedure.
g NaHCO3 = 45.00 molar mass NaHCO3 = 84.007 according to my net source.
moles NaHCO3 = (45.00/84.007) = 0.535669 and since you have 4 significant figures in the 45.00 g NaHCO3, this should be rounded to 0.5357 moles NaHCO3. Since there is 1 mole Na^+ in a mole of NaHCO3, the moles Na^+ = 0.5357 and that is the molarity. (22.99/83.99 is NOT 0.54--nor close to that).
c. N = M x #H atoms. N = 0.5357 x 1 = ??
d. 45.00 g NaHCO3 x (molar mass CaCO3/molar mass NaHCO3) = 45.00 g NaHCO3 x(100.087/84.007) = ??g CaCO3/L
Change that to mg CaCO3/L.
b you have close to the correct answer but not the correct procedure.
g NaHCO3 = 45.00 molar mass NaHCO3 = 84.007 according to my net source.
moles NaHCO3 = (45.00/84.007) = 0.535669 and since you have 4 significant figures in the 45.00 g NaHCO3, this should be rounded to 0.5357 moles NaHCO3. Since there is 1 mole Na^+ in a mole of NaHCO3, the moles Na^+ = 0.5357 and that is the molarity. (22.99/83.99 is NOT 0.54--nor close to that).
c. N = M x #H atoms. N = 0.5357 x 1 = ??
d. 45.00 g NaHCO3 x (molar mass CaCO3/molar mass NaHCO3) = 45.00 g NaHCO3 x(100.087/84.007) = ??g CaCO3/L
Change that to mg CaCO3/L.
In my book the anser for D is 2.68 x 10^4 mg/L as CaCO3, my first thought was to do what you did but i cant get to the answer in the book, am I missing something very obvious here? Im getting like 5.4 x 10^4
why is the molar mass 50 not 100 for the caco3?
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