titrate problems; please answer 4 and 5

Question

A solution of 0.04M hydroiodic acid is utilized to titrate 250mL of 7.0x10^-2M potassium lactate.

1. What is the pH of the initial potassium lactate solution? pH=8.3

1. What volume of HI needs to be added to the solution potassium lactate to reach the equivalence point? 437mL needed
2. What is the pH of the solution at the equivalence point? pH=2.7
3. The midpoint of the titration is where the amount of HI solution added is half the volume needed to reach the equivalence point of the titration. What is the pH in the solution at the midpoint of the titration? show work.
4. What is the pH of the solution once 100mL of the hydroiodic acid solution has been added to the flask? show work. 

Answer 1

1. 
   

We are told in the original problem that the potassium lactate has a concentration of 7.0x10-2M. The potassium is not chemically active as an acid nor base. But the lactate anion, C3H5O3- is the conjugate base of lactic acid, CH3H5O3H. From a table online that lists acids and their pKas, we know that lactic acid has a pka of 3.86. Working backward, we can use this known information to learn about our base. 

(pKa of the acid) + (pKb of the conjugate base)= 14 (always)

3.86 + pKb = 14

pKb = 10.14 = - log (Kb) 

Kb = 7.24 x 10^-11 (there is no unit) 

The Kb is the equilibrium constant for the reaction of the base lactate combining with water to produce lactic acid, and a hydroxide anion (OH-).

Here is the reaction: Lactate- + H2O → lactic acid + OH- 

CH3H5O3- + H2O → CH3H5O3H + OH- 

The formula for any equilibrium constant is K= (product of products)/ (product of reactants) with units of concentration in molarity or pressure (for gasses only). Pure substances like H2O are not included in the formula.

The pH is directly related (as we'll see later) to the concentration of OH- which is one of the products.

K= (product of products)/ (product of reactants)

7.24 x 10^-11 = (OH-)(lactic acid) / (lactate)

The given 7.0x10-2 M lactate is the total concentration for both forms of that species, the acid and the base, lactate and lactic acid. However, for a weak base like lactate, we can assume that the amount of lactate that has been changed to lactic acid is so low (by a magnitude 11x less than 10, given by the kb) that doing the calculation wouldn't change the answer by very much at all. Going back to the chemical equation, we can see that for every 1 mol of lactate, 1 mol of lactic acid and 1 mol of OH- is made. In other words, the amount of lactic acid and OH- will always be the same as each other here, so we can use one variable "x" to represent them both.

7.24 x 10^-11 = (OH-)(Lactic acid) / (lactate)

7.24 x 10^-11 = (x)(x) / (7.0x10-2 M)

7.24 x 10^-11 = x^2 / (7.0x10-2 M) 

Solving for x, x = 2.25 x 10^-6 M where x = the concentration of OH-

To get from the concentration of the OH- to the pH we need to know that the "p" simply means "the negative log of the concentration" and that pOH + pH = 14 always.

pOH = the negative log of the concentration of the OH = - log (2.25 x 10^-6 M) = 5.65 

pOH + pH = 14 so pH = 8.35 

1. 
   

The meaning of the equivalence point is that an equal amount (in moles) of acid and base will have been added. We have the molarity and volume for the base, lactate, provided by the problem. So the first step is to use that information to calculate the original moles of base. 

250mL of 7.0x10-2M potassium lactate.

7.0x10-2 mol/L * (250 mL) * (1 L / 1000 mL) = 0.0175 mol base

And we know that the moles of base will equal the moles of acid. We have the concentration of the acid provided by the problem, and so we can solve for the volume. 

[ (0.0175 mol) / (0.04 mol/L) ] * (1000 mL/ 1 L) = 438 mL acid

1. 3. 

For the titration, the following reaction occurs: 

Acid + base → conjugate base + conjugate acid

The proton (H+) has moved from the acid to the base, forming the conjugate species

HI + CH3H5O3- → CH3H5O3H + I- 

For the equivalence point, the same amount (in moles) of the original acid and base have been added. So once the reaction is complete, the only thing that remains are the products. 

HI is considered a strong acid (there are 6 strong acids universally recognized). The conjugate base of a strong acid (here that is the I-) is not considered chemical active as an acid nor a base, so it can be ignored. 

Therefore the pH will only be impacted by the CH3H5O3H, the conjugate acid of lactate, called lactic acid. From part 1, we already know that the pKa of lactic acid is 3.86. 

pKa = - log (Ka). Ka = 10^(-pKa) = 1.38 x 10^-4 (there are no units) 

To find the concentration of lactic acid, we need to consider the number of moles of lactate (which we already calculated in part 2 as 0.0175 mol) and the volume, which we know because of the original volume of 250 mL base and the added 438 mL of the acid, calculated in part 2. 

[ 0.0175 mol / (250 mL + 438 mL) ] * (1000 mL/ 1 L) = 0.0254 M lactic acid 

The only reactive species in the solution is the lactic acid. It reacts like any other acid. 

Chemical reaction: lactic acid + H2O → lactate- + H+ 

The formula for any equilibrium constant is K= (product of products)/ (product of reactants) with units of concentration in molarity or pressure (for gasses only). Pure substances like H2O are not included in the formula.

K= (product of products)/ (product of reactants)

K = (H+)(lactate) / (lactic acid)

Going back to the chemical equation, we can see that for every 1 mol of lactic acid, 1 mol of lactate and 1 mol of H+ is made. In other words, the amount of lactate and H+ will always be the same as each other here, so we can use one variable "x" to represent them both.

1.38 x 10^-4 (calculated above) = (H+)(lactate) / (0.0254 M) 

1.38 x 10^-4 = (x^2) / (0.0254 M) 

Solving for x, x = 3.51 x 10^-6 M where x = the concentration of H+

pH = - log (H+) = - log (3.51 x 10^-6 M) = 5.46