1. In what total volume (mL) must 331.3 mg of NaOH be dissolved in order to make

Question

1. In what total volume (mL) must 331.3 mg of NaOH be dissolved in order to make a solution with a molar analytical concentration of 0.353 M?

2. What mass in grams of sodium carbonate, Na2CO3, must be used to make 1.94 L of a solution in which the equilibrium concentration of Na+ is 0.835 M?

3. What mass in grams of barium chloride monohydrate, BaCl2•H2O, must be used to make 2.83 L of a solution in which the equilibrium concentration of Cl- is 0.873 M?

4. For the following dissociation of acetic acid, K = 1.76 x 10-5: CH3COOH CH3COO- + H+

Sodium acetate completely dissociates in solution according to the following reaction:NaCH3COO Na+ + CH3COO-

What volume in mL of 4.85 M acetic acid must be added to 99.95 mg of sodium acetate, such that when diluted with water to a final volume of 566.2 mL, the equilibrium concentration of H+ is 5.97 x 10-5 M?

Explanation / Answer

1. molarity = moles/L of solution

moles = g of solute/molar mass

molar mass of NaOH = 40 g/mol

So, To get a molarity of 0.353 M,

Volume = 0.3313/(0.353 x 40) = 0.0235 L = 23.5 mL is required

2. 1 mole of Na2CO3 has 2 moles of Na+

that is, 106 g of Na2CO3 has 80 g of Na+

[Na+] = 0.835 M

So, [Na2CO3] = 1/2 x 0.835 M = 0.4175 M

mass of Na2CO3 required would be = 0.4175 M x 106 g/mol x 1.94 L = 85.855 g

3. 1 mole of BaCl2.2H2O has 2 moles of Cl-

that is, 226.245 g of BaCl2 has 70 g of Cl-

[Cl-] = 0.873 M

So, [BaCl2.H2O] = 1/2 x 0.873 = 0.4365 mole

mass of BaCl2.H2O required would be = 0.4365 mole/L x 226.245 g/mole x 2.83 L = 279.45 g

4. [H+] = 5.97 x 10^-5 M

[CH3COOH] = 4.85 M

K = 1.76 x 10^-5 = [CH3COO-][H+]/[CH3COOH] = [CH3COO-](5.97 x 10^-5)/(4.85)

[CH3COO-] = 1.43 M

moles of sodium acetate = 0.09995/82.0343 = 1.22 x 10^-3 mole

Molarity of sodium actetate in final solution = 1.22 x 10^-3/0.5662 = 2.155 x 10^-3 M

So, to get 2.155 x 10^-3 M we would need = 2.155 x 10^-3 x 0.5662/1.43 = 8.53 x 10^-4 L = 0.853 mL of acetic acid is required

thus we would need 0.853 mL of 4.85 M acetic acid to get final H+ as 5.97 x 10^-5 M  

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