1. In the laboratory you dissolve 12.6 g of barium acetate in a volumetric flask
ID: 563073 • Letter: 1
Question
1. In the laboratory you dissolve 12.6 g of barium acetate in a volumetric flask and add water to a total volume of 250 mL.
What is the molarity of the solution? ____ M.
What is the concentration of the barium cation? ____ M.
What is the concentration of the acetate anion? ____ M.
2. In the laboratory you dissolve 13.7 g of potassium fluoride in a volumetric flask and add water to a total volume of 375 mL.
What is the molarity of the solution? ____ M.
What is the concentration of the potassium cation? ____ M.
What is the concentration of the fluoride anion? ____ M.
3. In the laboratory you dissolve 20.8 g of iron(II) acetate in a volumetric flask and add water to a total volume of 250 mL.
What is the molarity of the solution? ___ M.
What is the concentration of the iron(II) cation? ___ M.
What is the concentration of the acetate anion? ____ M.
Explanation / Answer
1. molarity of solution = moles of solute/ volume of solution (in lit)
given mass of barium acetate = 12.6 gram; molar mass of C4H6O4Ba (barium acetate) = 255.43 g/mol
moles of barium acetate= 12.6/255.43 = 0.0493 mol
molarity = 0.0493/0.25 =0.1973 mol/lit orv (M)
concentration of barium cation = 0.1973 M ( because each mole of barium acetate gives 1 mol of Ba2+)
concentration of acetate anion = 2x 0.1973 = 0.3946 M
2.mass of KF = 13.7 gram, molar mass of KF =58.09
moles of KF = 13.7/58.09 = 0.2358 mol
concentration of KF = 0.2358 mol/ 0.375 lit =0.6289 M
since each mol of KF gives one mol of K+ and on mol of F-
concentration of K+ = 0.6289 M
concentration of F- = 0.6289 M
3.mass of Iron(II) acetate = 173.93 g/mol
=> moles of Iron(II) acetate = 20.8g/ 173.93 g/mol = 0.1195 mol
concentration of Iron(II) acetate = 0.1195/0.25 =0.4783 M
since Iron(II) acetate is Fe(C2H3O2)2
therefore concentration of iron 2+ = concentration of Iron(II) acetate = 0.4783 M
concentration of acetate ion = 2xconcentration of Iron(II) acetate = 2x 0.4783 =0.9567 M
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