Question

The recommended dose of quinine is 0.800 g/kg. If we assume the density is 1 g/ml what is the ph of this solution? the molar mass of quinine is 324.412. If 100 ml of this solution is titrated with a solution of 0.010 M HCL how many ml of HCL are required to reach the first equivalence point? what is the ph there? What is the PH when 32 ML of HCL have been added? what is the ph at the second equivalence point? what species of quinine is predominant at a ph of 2.5(in the stomach)? and 7.4 (in the blood)

Kb1= 1.0x10^-6

Kb2= 1.58x10^-10

Answer 1

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Assuming density of 1 g/ml concentration of the solution will be 0.800 g/ L. We calculate the molar concentration from this concentration:

M = 0.800 g/ 324.412 gmol^-1 /1L = 2.466 x 10^-3 M

No we write the hydrolysis reaction for a base:

Since Quinine has two values of Kb, we pick the larger (or bigger) Kb because this will be the first ionization for the base. At equilibrium we can find the concentrations of each of them and the pH by assuming that the weak base (represented by B:) dissociates in an X amount, so there will be X mol of Conjugated acid (BH⁺) and X mol of OH^-

Solving for X we obtain:

Since the base is a very weak base we can simplify the equation by saying that X is a very small portion son C₀ – X is almost equal to C₀. In this case we can find the X with a more simple expression:

ll be X mol of Conjugated acid (BH⁺) and X mol of OH^-

Remember that Kb is 1 x 10^-6

The answer for X = 4.965 x 10^-5

The pOH is –logX = 4.304 Which means that pH = 14 – pOH = 14 – 4.304 = 12.696

Second question. Titration of 100 mL of the Quinine solution

Titration requires that we use exactly the same number of mol of acid that the base that we are valorizing. So we calculate the number of mol of quinine:

n = 0.1 L x 2.466 x 10^-3 M = 0.2466 x 10^-3 mol

In equivalent point the number of mol of base and acid are the same. We know [HCl] = 0.01 M so the used volume will be:

V[HCl] = 0.2466 x 10^-3 mol / 0.01 M = 24.66 ml

Third question pH. In the equivalent point we have the conjugated acid of quinine, which is a weak acid and hydrolyzes according to:

We can solve this equation exactly the same way we did for the first question. To find Ka for this reaction we use the relation: Ka x Kb = 10^-14. This makes Ka for the reaction 1 x 10^-8

C₀ = 0.2466 x 10^-3 mol / 0.1246 mL = 1.979 x 10^-3 M

With this Ka value pH will be: 5.35

Question 4: pH after adding 32 mL of HCl

Once the titration passes the first equivalent point, HCl starts reacting with the second basic function (Kb = 1.58 x 10^-10). The number of mol starting will be the same we already have so we start with 0.2466 x 10^-3 mol of the substrate which happens to be the conjugated base of quinine. Since we are starting with the same amount and titrating with the same acid we expect to use double of the amount for the second equivalent point. That will be around 49.32 mL. So this means that before that point we are on a buffer zone and the pH should change very little before we get close to the equivalent point. Since 32 mL is far from that the pH should be very close to 5.35

Question 5: pH at the second equivalent point. Calculate this pH exactly as in question 3 but using Kb = 1.58 x 10^-10 this makes Ka for the reaction to be 6.33 x 10^-5

and

C₀ = 0.2466 x 10^-3 mol / 0.1493 mL = 1.652 x 10^-3 M

Questions 6 and 7:

These two questions solve with Henderson-Hasselbach equation:

For both pH we have to use the first pKa because pH is lower than 10 in both cases:

pH 2.5 (stomach):

[B]/[BH⁺] = 10^-5.5

This number is extremely small, which means that the protonated form is much higher that the not protonated form.

pH 7.4 (blood)

Using the same equation we get [B]/[BH⁺] = 10^-0.6

This means that at blood’s pH there are almost equal amounts of protonated and not protonated form.

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