Question

Explain how you would use a spreadsheet such as Figure 8-9 to find the and concentrations of all species if you know the values of the equilibrium constants and the formal concentration $\mathit{F}$

Short answer

The equilibrium constants and the formal concentration F is

$\mathrm{F}=\left[{\mathrm{AI}}^{3+}\right]+\left[{\mathrm{AIOH}}^{2+}\right]+\left[\mathrm{AL}{{\left(\mathrm{OH}\right)}_2}^{+}\right]+2\left[{\mathrm{AL}}_2{{\left(\mathrm{OH}\right)}_2}^{+4}\right]+\left[\mathrm{AL}{{\left(\mathrm{OH}\right)}^{-}}_4\right]+\left[{\mathrm{AI}}_3{{\left(\mathrm{OH}\right)}^{+5}}_4\right]$

Step-by-step solution

Define the equilibrium constant.

The ratio of the concentration of products to the concentration of reactants, each raised to their respective stoichiometric coefficients, is the equilibrium constant of concentration (denoted by K c) of a chemical reaction at equilibrium.

Find the ph of the concentration.

If all of the values for the equilibrium constant and formal concentration are known, the p^H of alclo4 and the concentration can be found by solving the 10 equations with their unknowns.

The values are calculated when the appropriate constants and formulas are entered in the spreadsheet cells.

$3\left[{\mathrm{AI}}^{3+}\right]2+\left[{\mathrm{AIOH}}^{2+}\right]+\left[\mathrm{AL}{{\left(\mathrm{OH}\right)}_2}^{+}\right]+4\left[{\mathrm{AL}}_2{{\left(\mathrm{OH}\right)}_2}^{+4}\right]+5\left[\mathrm{AL}{{\left(\mathrm{OH}\right)}^{5+}}_4\right]+\left[{\mathrm{H}}^{+}\right]$

$=\left[\mathrm{AI}{{\left(\mathrm{OH}\right)}_4}^{-}\right]+\left[0{\mathrm{H}}^{‐}\right]+\left[{{\mathrm{CIO}}_4}^{-}\right]$

Then the mass $3\mathrm{F}=\left[{{\mathrm{CIO}}_4}^{-}\right]$

$\mathrm{F}=\left[{\mathrm{AI}}^{3+}\right]+\left[{\mathrm{AIOH}}^{2+}\right]+\left[\mathrm{AL}{{\left(\mathrm{OH}\right)}_2}^{+}\right]+2\left[{\mathrm{AL}}_2{{\left(\mathrm{OH}\right)}_2}^{+4}\right]+\left[\mathrm{AL}{{\left(\mathrm{OH}\right)}^{-}}_3\right]+\left[{\mathrm{AI}}_3{{\left(\mathrm{OH}\right)}^{+5}}_4\right]+3\left[{\mathrm{AI}}_3{{\left(\mathrm{OH}\right)}^{+5}}_4\right]$