How Do You Calculate OH- Concentration in Benzoic Acid to Benzoate Conversion?

[mountain]

my book gives an example of how to find the concentration of OH- for transforming 99% benzoic acid to benzoate anion. it is difficult to understand how they get the answer, since there are no specified concentrations for the benzoic acid and benzoate anion, besides the way they write it is also difficult to understand. this is how they write:

the ability to separate a strong from weak acids depends on the acidity constants of the acids and the basicity constants of the bases as folllows. in the first equation, consider the ionization of benzoic acid, which has an equilibrium constant Ka of 6.8x10-5. the conversion of benzoic acid to the benzoate anion in the fourth equation is governed by the equilibrium constant K (Eq.5), obtained by the combining the third and fourth equations.

C6H5COOH + H2O -> <- C6H5COO- + H3O+ (Eq. 1)

Ka= [C6H5COO-][H3O+]/[C6H5COOH]=6.8x10-5, pKa=4.17 (Eq. 2)

Kw=[H3O+][OH- ]=10-14 (Eq. 3)

C6H5COOH + OH- -> <- C6H5COO- + H2O (Eq. 4)

K=[C6H5COO-]/[C6H5COOH][OH- ]=Ka/Kw=6.8x10-5/10-14=6.8x109 (Eq. 5)

if 99% of the benzoic acid is converted to C6H5COO- :

[C6H5COO-]/[C6H5COOH]=99/1 (Eq. 6)

then from Eq. 5 the hydroxide ion concentration would need to be 6.8x10-7M.

my question, how do they get [OH- ]= 6.8x10-7M ? :grumpy:

hope for any ideas and guidance!

thanks for helping!

 

[Navin A S]

The answer of 6.8x10-7M comes from combining the equations and solving for [OH- ]. Using equation 5, we can rearrange it to [OH- ] = Ka/Kw x [C6H5COOH]/[C6H5COO-]. Since Ka is given as 6.8x10-5 and Kw is given as 10-14, then the equation simplifies to [OH- ] = 6.8x10-5/10-14 x [C6H5COOH]/[C6H5COO-]. From equation 6, we know that [C6H5COO-]/[C6H5COOH] = 99/1, so the equation becomes [OH- ] = 6.8x10-5/10-14 x 1/99, which gives us [OH- ] = 6.8x10-7M.

 

[ravenprp]

The equations provided in the book may be confusing, but essentially they are showing the equilibrium reactions that occur when benzoic acid is converted to benzoate anion. The first equation (Eq. 1) shows the ionization of benzoic acid, where it loses a hydrogen ion (H+) to form the benzoate anion (C6H5COO-). This reaction is governed by the equilibrium constant Ka, which is equal to the concentration of the products (C6H5COO- and H3O+) divided by the concentration of the reactant (C6H5COOH). The value of Ka for benzoic acid is given as 6.8x10-5 in the second equation (Eq. 2), which is also known as the acidity constant. The pKa value (equivalent to -logKa) is also given as 4.17, which is a measure of the strength of the acid.

The third equation (Eq. 3) shows the equilibrium constant for the autoionization of water (Kw), which is equal to the concentration of hydrogen ions (H3O+) multiplied by the concentration of hydroxide ions (OH-). In pure water, the concentration of hydrogen ions and hydroxide ions is equal, so Kw is equal to 10-14. This equation is used to find the concentration of OH- in the solution.

The fourth equation (Eq. 4) shows the reaction that occurs when a base (in this case, hydroxide ions) is added to benzoic acid. This reaction is also governed by an equilibrium constant, which is equal to the concentration of the products (C6H5COO- and H2O) divided by the concentration of the reactants (C6H5COOH and OH-). This equilibrium constant is denoted as K in the fifth equation (Eq. 5), which is obtained by combining the third and fourth equations. Since the concentration of OH- is unknown, it is represented as [OH-] in this equation.

The last equation (Eq. 6) shows the ratio of the concentration of benzoate anion to benzoic acid when 99% of the benzoic acid is converted. This ratio is equal to 99/1, since 99% of the benzoic acid is converted to benzoate anion. From Eq.