Determine the molar solubility of Al(OH) 3 in a solution containing 0.0500 M AlC

Question

Determine the molar solubility of Al(OH)3 in a solution containing 0.0500 M AlCl3. Ksp (Al(OH)3) = 1.3 × 10-33.

Determine the molar solubility of MgCO3 in pure water. Ksp (MgCO3) = 6.82 × 10-6.

       If the pKa of HCHO2 is 3.74 and the pH of an HCHO2/NaCHO2 solution is 3.11, which of the following is TRUE?

       Answers:

       [HCHO2] << [NaCHO2]

       [HCHO2] < [NaCHO2]

       [HCHO2] = [NaCHO2]

       [HCHO2] > [NaCHO2]

       It is not possible to make a buffer of this pH from HCHO2 and NaCHO2.

Explanation / Answer

Determine the molar solubility of Al(OH)3 in a solution containing 0.0500 M AlCl3. Ksp (Al(OH)3) = 1.3 × 10-33.

Given that;

Ksp (Al(OH)3) = 1.3 × 10-33.

[AlCl3] = 0.0500 M

One moles of AlCl3 gives one moles Al3+ and three moles Cl-

AlCl3 --- > Al+3 + 3Cl-

[ Al+3] = 0.0500 M

Al(OH)3(s) --- > Al+3 + 3OH-

One moles of Al(OH)3 gives one moles Al3+ and three moles OH-

The solubility product of Al(OH)3 is given as follows:

Ksp = [Al+3][OH-]^3

assume that the x is the molar solubility of Al(OH)3
[Al+3] = 0.0500 +x
[OH-] = 3x

(0.0500 + x)(3x)^3 = 1.3 x 10^-33

Here 0.0500 + x = 0.0500 due to small value of x.

Therefore;

0.0500(27x^3) = 1.3 x 10^-33
1.35x^3 = 1.3 x 10^-33
x^3 = 0.96 x 10^-33
x = 9.86 x 10^-12
         

Determine the molar solubility of MgCO3 in pure water. Ksp (MgCO3) = 6.82 × 10-6.

Given that; Ksp (MgCO3) = 6.82 × 10-6

   Assume that the x M is the molar solubility of MgCO3 in pure water

Mg -- >Mg2+ + CO3^2-

Ksp = [Mg2+][CO32-]= (x)(x)

6.82 × 10-6 = x ^2

X=2.61 x 10^-3 M          

If the pKa of HCHO2 is 3.74 and the pH of an HCHO2/NaCHO2 solution is 3.11, which of the following is TRUE?

       Answers:

       [HCHO2] << [NaCHO2]

       [HCHO2] < [NaCHO2]

       [HCHO2] = [NaCHO2]

       [HCHO2] > [NaCHO2]

First we determine the ratio of        [HCHO2] and [NaCHO2] as follows:

pH = pKa + log [salt]/[acid]

pH = pKa + log [NaCHO2]/[HCHO2]

3.11= 3.74+ log [NaCHO2]/[HCHO2]

log [NaCHO2]/[HCHO2]= -0.63

[NaCHO2]/[HCHO2]=10^ -0.63

[NaCHO2]/[HCHO2] = 0.23
hence [HCHO2] > [NaCHO2] is correct.

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