1) A mixture of equal volumes of equimolar solutions of acetic acid and sodium a

Question

1) A mixture of equal volumes of equimolar solutions of acetic acid and sodium acetate is made in 1 L flask. Calculate the pH of the solution.

2) A sample of 10.0 mL solution of 0. 100 M acetic acid is titrated with 0.100 M NaOH solution. Calculate the pH of the solution when exactly 5.00 mL of the base has been added.

3) Calculate the solubility in molarity of Cu(OH)2, (a) In pure water and in (b) 0.010M Cu(NO3)2 solution. The Ksp for Cu(OH)2 is 2.2O x 10^-20.

4) (a) Calculate the value for Ksp for CaSO4 that has the solubility of 0.67 g/L in water.

(b) If we mix 50.0 mL of 0.100 M Na2SO4 with 5.00 mL 0f 0.100 M BaCl2, will there be any precipitation of BaSO4? Look up for Ksp value for BaSO4.

I'm looking for good explanation, please be as thorough as possible.

Explanation / Answer

2.

a) Calculate moles of acid and base in solution before reaction:

CH3COOH: 1 x 10-3 mol
NaOH: (0.00500 L) (0.100 mol/L) = 5 x 10-4 mol

b.) Determine amounts of acid and acetate ion after reaction:

CH3COOH: 1 x 10-3 mol - 5 x 10-4 mol = 5 x 10-4 mol
CH3COONa: 5 x 10-4 mol

c) Use Henderson-Hasselbalch equation to determine pH of (now) buffered solution:

pH = 4.752 + log (5.0 x 10-4 / 5 x 10-4)

pH = 4.752 + 0

pH = 4.752

4.

First change 0.67g/L into molarity

(0.67g/L)(1mol/136.15g)= 4.9 x 10-3 M

Then write a chemical equation for the dissolution of CaSO4

CaSO4 <--> Ca2+ + SO4 2-

setting up an equilibrium expression gives us
[Ca2+][SO4 2-] = ksp
since the concentrations of Ca and SO4 are the same, we end up with

(4.9 x 10-3)2 = 2.4 x 10-5

so the ksp of CaSO4 is 2.4 x 10-5..

3.

a.Ca[OH]2 >>>>> Ca+2 + 2[OH]-1

So your equation is:

Ksp = [x][2x]2 = 2.2 x 10-20

[x][2x]^2 = 4x^3 = 2.2 x 10-20

x = [(2.2 x 10-20)/4]1/3 =1.76 x 10-7.

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