Lactic acid C3H6O3 is found in sour milk where it is produced by the action of lactobacilli in lactose or the sugar in milk. The pH of a 0.045 M solution of lactic acid was determined using a pH probe and found to be 2.63. a. Calculate the equilibrium constant for this acid. b. Had you not been given the pH of the acid and you had to measure it yourself, how would the method in part 2 be applied to the determination of Ka? Would you expect an improvement in the accuracy of your result with the application of the method of this experiment? Explain why or why not.

see explanation below

Explanation:

In this case, we need to write the overall reaction:

HC₃H₅O₃ + H₂O <> C₃H₅O₃⁻ + H₃O⁺

The lactic acid is a weak acid, so, when it dissociates in it's ions, part of the acid is dissociated. This depends of it's Ka to know which quantity was dissociated.

To calculate Ka, let's write an ICE chart first:

     HC₃H₅O₃ + H₂O <> C₃H₅O₃⁻ + H₃O⁺      Ka = ?

i)       0.045                                  0            0

c)          -y                                     +y           +y

e)   0.045 - y                                 y              y

Writting the Ka expression we have:

Ka = [C₃H₅O₃⁻] [H₃O⁺] / [HC₃H₅O₃]

Now, to calculate Ka we need the values of [C₃H₅O₃⁻] and [H₃O⁺] in equilibrium. Fortunately, we have the value of the pH, which is 2.63 and with this we can get the value of [H₃O⁺] and then, the value of y. With that value, we replace it in the Ka expression to calculate Ka:

[H₃O⁺] = 10^(-pH)

[H₃O⁺] = 10^(-2.63)

[H₃O⁺] = [C₃H₅O₃⁻] = x = 2.34x10⁻³ M

Now, let's replace this value in the Ka expression:

Ka = (2.34x10⁻³)² / (0.045 - 2.34x10⁻³)

Ka = 1.28x10⁻⁴

b) Now, let's calculate the pH with the obtained value of Ka. We will use the same expression of Ka so:

1.28x10⁻⁴ = y² / (0.045-y)    

1.28x10⁻⁴ (0.045 - y) = y²

5.76x10⁻⁶ - 1.28*10⁻⁴y = y²

y² + 1.28x10⁻⁴y - 5.76x10⁻⁶ = 0

From here, we'll use the quadratic equation general formula, for solving y:

y = -1.28x10⁻⁴ ±√(1.28x10⁻⁴)² + 4 * 1 * 5.76x10⁻⁶ / 2

y =  -1.28x10⁻⁴ ±√2.31x10⁻⁵ / 2

y = -1.28x10⁻⁴ ± 4.8x10⁻³ / 2

y₁ = 2.34x10⁻³ M

y₂ = -2.464x10⁻³ M

Now, as y₁ is positive this is the value that we will take.

This value would be the [H₃O⁺] in equilibrium.

The value of pH would be:

pH = -log[H₃O⁺]

pH = -log(2.34x10⁻³)

pH = 2.631

According to this value of pH we can actually expect an inprovement in the accuracy, basically because we obtain a value with more significant figures, and this are relationed with accuracy.

its not a choice. you dont choose if you want to be gay or straight.

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.