4. [34] L-leucine hydrochloride, [(CH,),CHCH,CHONH)COOHI\'C may be abbreviated a

Question

4. [34] L-leucine hydrochloride, [(CH,),CHCH,CHONH)COOHI'C may be abbreviated as Ha used as the dietary supplement for the essential amino acid leucine. As you can see this salt cationic acid with pKa-2.36 and pK,2 -9.60. A 50.0 mL of a 0.100 M solution of leucine h is titrated with 0.500 M NaOH solution. (a) Calculate the initial pH of the 0.100 M leucine hydrochloride solution? (b) As the titration proceeds, what volume of 0.500 M NaOH must be added to achieve a pH c (c) Calculate the maximum concentration of the leucine hydrochloride species at the first equ point in the titration and the pH of the solution? (d) The titration is continued until the second equivalence point is reached. Calculate the pH solution and select an appropriate indicator from the table below that can be used to signal the of the titration. Indicator Color Changes idic pH range of color T Color of Aci pKI Color Indicator for bl? bl? re ell formchange yellow8.0-9.6 8.9 9 thymol blue thymolphthalein alizarin yellow R Indigo Carmen ellow ellow blue 9.3- 10.5 10.1 12.0 11.4-13.0 9.9 11.2 12.2

Explanation / Answer

(a) The first dissociation that takes place is:

H2A + H2O = HA- + H3O + Ka1 = 4.37 x10 ^ -2

Using the equilibrium relationship we have:

Ka = [HA -] * [H3O +] / [H2A] = x ^ 2 / (0.1-x)

We clear the x and we have:

x = 0.019 = [HA-] = [H3O +]

pH = -Log (0.019) = 1.72

(b) It is necessary to know the pH required by the exercise.

(c) First we must calculate the volume of the first point of equivalence, for this we use the neutralization relation:

Ca * Va = Cb * Vb

We clear the necessary base volume and we have:

Vb = Ca * Va / Cb = 0.019 M * 50 ml / 0.5 M = 1.9 mL

We have the total volume of solution: Vt = 50 ml + 1.9 ml = 51.9 ml = 0.0519 L

moles of H2A that did not dissociate:

(0.1 - 0.019) mol / L * 50 mL / 1000 ml / L = 0.00405 mol H2A

Concentration:

[H2A] = 0.00405 mol / 0.0519 L = 0.078 M

For the second dissociation, the following is fulfilled:

HA- + H2O = A-2 + H3O + Ka2 = 2.51 * 10 ^ -10

balance ratio is left:

Ka = x ^ 2 / 0.019 - x

clearing x = 2.18 * 10 ^ -6 = [H3O +]

pH = - Log (2.18 * 10 ^ -6) = 5.66

(d) The second point of equivalence is met to the 3.8 ml of NaOH added, to calculate the pH at this point, we will find the remaining moles of NaOH:

Moles Total NaOH = 0.5 mol / L * 3.8 ml / 1000 ml / L = 0.0019 mol NaOH added

Moles H2A neutralized:

1st Point: 0.019 mol / L * 50 mL / 1000 ml / L = 0.00095 mol

2nd Point: 2.18 * 10 ^ -6 mol / L * 51.9 ml / 1000 ml / L = 0.000000113 mol

Total H2A mole = 9.50113 * 10 ^ -4

Remaining moles of NaOH = 0.0019 - 9.50113 * 10 ^ -4 = 9.5 * 10 ^ -4

pOH = -Log (9.5 * 10 ^ -4) = 3.02

pH = 14- pOH = 14- 3.02 = 10.98

The Yellow alizarin indicator must be chosen because of its proximity to the final valuation point.

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