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What is the pH of a 0.097 M solution of lithium lactate? (Ka for lactic acid is 1.4x10-4)

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Answer: To calculate the pH of a solution of lithium lactate, we first need to understand the dissociation of lactic acid (HC3H5O3) in water. Lactic acid is a weak acid that partially dissociates into its conjugate base (lactate, C3H5O3-) and hydrogen ions (H+).

The balanced chemical equation for the dissociation of lactic acid is:

HC3H5O3 (lactic acid) ⇌ C3H5O3- (lactate) + H+ (hydrogen ion)

The equilibrium expression for this dissociation reaction is given by:

Ka = [C3H5O3-][H+] / [HC3H5O3]

where Ka is the acid dissociation constant for lactic acid, [C3H5O3-] is the concentration of lactate, [H+] is the concentration of hydrogen ions, and [HC3H5O3] is the concentration of lactic acid.

We are given the Ka value for lactic acid, which is 1.4 x 10^-4. The concentration of lactic acid (HC3H5O3) in the solution is not given directly, but we can assume that it dissociates to a negligible extent compared to the initial concentration of lithium lactate (LiC3H5O3), which is 0.097 M. Therefore, we can approximate the concentration of lactic acid ([HC3H5O3]) as 0.097 M.

Now, let's calculate the concentration of lactate ([C3H5O3-]) and hydrogen ions ([H+]) using the equilibrium expression and the known Ka value:

Ka = [C3H5O3-][H+] / [HC3H5O3]

1.4 x 10^-4 = [C3H5O3-][H+] / 0.097

[C3H5O3-] = [H+] = (1.4 x 10^-4) * 0.097

[C3H5O3-] ≈ [H+] ≈ 1.36 x 10^-5 M

Now, to calculate the pH, we need to find the negative logarithm of the hydrogen ion concentration:

pH = -log[H+]

pH ≈ -log(1.36 x 10^-5)

pH ≈ -(-4.87)

pH ≈ 4.87

Therefore, the pH of a 0.097 M solution of lithium lactate is approximately 4.87.

User Andrew Ferrier
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