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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

  • chemistry -

    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.

  • chemistry -

    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

  • chemistry -

    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.

  • chemistry -

    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

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