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Calcium propionate [Ca(CH₃CH₂COO)₂; calcium propanoate] is a mold inhibitor used in food, tobacco, and pharmaceuticals. (b) Use Appendix C to find the resulting pH when 8.75 g of Ca(CH₃CH₂COO)₂ dissolves in enough water to give 0.500 L of solution.

Respuesta :

The resulting pH when 8.75 g of Ca(CH₃CH₂COO)₂ dissolves in enough water to give 0.500 L of solution.

The pH of the solution is = 9.08

What is pH value:

Historically denoting "potential of hydrogen," is a scale used to specify the acidity or basicity of an aqueous solution. Acidic solutions have lower pH values than basic or alkaline solutions.

According to the given information:

To determine the initial concentration of propionate anions in solution, use the following formula:

                         [tex]\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COO}^{-}\right]=\frac{\frac{17.5 \mathrm{~g}}{186.22 \mathrm{gmol}^{-1}}}{0.500 \mathrm{~L}}=0.188 \mathrm{M}[/tex]

At equilibrium: The equilibrium equation becomes

                         [tex]K b=\frac{x^2}{0.188-x}[/tex]

Using the small x approximation to solve for x:

                        [tex]\begin{gathered}K b=\frac{1.0 \times 10^{-14}}{1.3 \times 10^{-5}} \\K b=7.69 \times 10^{-10} \\7.69 \times 10^{-10}=\frac{x^2}{0.188} \\x=\sqrt{\left(7.69 \times 10^{-10}\right)(0.188)} \\x=\left[O H^{-}\right]=1.20 \times 10^{-5}\end{gathered}[/tex]

Verifying the approximation:

                       [tex]\frac{1.20 \times 10^{-5}}{0.188} \times 100 \%=6.38 \times 10^{-3} \%[/tex]

The approximation is correct because the value is less than 5%.

Calculating pH:

                      [tex]p O H=-\log 1.20 \times 10^{-5}[/tex]

                      pOH=4.92

                      pH=14−4.92

                       pH=9.08

The pH of the solution is = 9.08

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