102.0 mL of a solution ofquinine contains 183.6 x 10-3 mol of quinine.?

Question

105.0 mL of a solution of quinine contains 126.0×103 mol of quinine. Quinine is monobasic with Kb = 7.90× 106

a, If this is titrated with 0.200 M HCl, what is the pH half-way to thestoichiometric point ?    b. For the titration in the above question,what is the pH at thestoichiometric point ? 105.0 mL of a solution of quinine contains 126.0×103 mol of quinine. Quinine is monobasic with Kb = 7.90× 106

a, If this is titrated with 0.200 M HCl, what is the pH half-way to thestoichiometric point ?    b. For the titration in the above question,what is the pH at thestoichiometric point ?    b. For the titration in the above question,what is the pH at thestoichiometric point ?

Explanation / Answer

I am a pretty rusty on this, so here is asimilar question with different numbers. The answer is veryin-depth so I think you will find it very helpful:
a) If this is titrated with 0.200 M HCl, whatis the pH half-way to the stoichiometric point ?
Quinine is monobasic with Kb = 7.90 x 10-6

b)What is the pH at the stoichiometric point ? a) If this is titrated with 0.200 M HCl, whatis the pH half-way to the stoichiometric point ?
Quinine is monobasic with Kb = 7.90 x 10-6

b)What is the pH at the stoichiometric point ? ______________________________________________________________ THE ANSWERs ARE :
-) pH1 = 8.9 (Half-Way) and
-) pH2 = 4.8 (Equivalence Point).

ABOUT TITRIMETRY
You executed an Acid-Base Titrimetric Determination.
As you wrote, Quinine acts as a base in its aqueous solutions,hence a STRONG ACIDIC STUFF as Hydrochloric Acid can react fastlyand complete by means of a
WELL-KNOWN STOICHIOMETRY (e.g. one mole of Quinine versus one moleof Hydrochloric Acid) : this reaction satisfy the TitrimetricRequests, it may pursue the Titrimetric Operations.
In the Titrimetric Determinations, the chemist define the"Equivalence Point", e.g. the Conditions where the he added aTitrimetric Reactive's Amount EQUAL THAN Analyte's One.

CALCULATIONs
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°...
You ask me the pH's Value of the Titrimetric Mixture at itsHALF-WAY STOICHIOMETRIC POINTs, e.g. the Half Way toward theEquivalence Conditions

(Quinine's Moles) = 2 * (HCl's Moles) =
= 2* (HCl's Molarity) * (HCl's Volume)
(HCl's Volume) = (1/2) * (Quinine's Moles) / (HCl's Molarity) =
= 0.5 * 0.1836 / 0.2000 = 0.4590 liter

Assuming the Volume Addiction's Property

(Quinine's Volume) + (HCl's Volume) = 0.1020 + 0.4590 =
= 0.5610 liter

I know the final volume which is related to the final concentrationof the chemical stuffs present in the mixture

(Final Quinine's Moles) =
= (Initial Quinine's Moles) - (Added HCl's Moles) =
= 0.1836 - (0.5 * 0.1836) = 0.0918 mol
(Final QuinineH+'s Moles) =
= (Initial Quinine's Moles) - (Final Quinine's Moles) =
= 0.1836 - 0.0918 = 0.0918 mol

I remember you that Quinine is the lonely Weak Electrolyte insteadHydrochloric Acid, so I back to the Acid-Base's Equilibrium ofQuinine's aqueous solutions

7.9E-6 = Kb = |QuinineH+|eq * |OH-|eq / |Quinine|eq

because this relation governs this System

|OH-|eq = Kb * |Quinine|eq / |QuinineH+|eq =
= 7.9E-6 * 0.0918 / 0.0918 = 7.9E-6 M
|H+|eq = Kw / |OH-|eq = 1.0E-14 / 7.9E-6 = 1.3E-9 M
that is pH1 = 8.9.
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°...
You ask me the pH's Value of the Titrimetric Mixture at itsEQUIVALENCE POINT, e.g. the Equivalence Conditions

(Quinine's Moles) = (HCl's Moles) =
= (HCl's Molarity) * (HCl's Volume)
(HCl's Volume) = (Quinine's Moles) / (HCl's Molarity) =
= 0.1836 / 0.2000 = 0.9180 liter

Assuming the Volume Addiction's Property

(Quinine's Volume) + (HCl's Volume) = 0.1020 + 0.9180 =
= 1.0200 liter

I overcome to the Final Concentration of the Stuff

(Final Stuff Concentration) =
= (Stuff's Amount) / (Final Volume) = 0.1836 / 1.0200 =
= 0.1800 M

HOW DO I CALCULATE THE pH's VALUE RELATED TO THISCONCENTRATION?
I write the mathematical expression of the scientific relationscharacterizing the ions present in the mixture :

i) 7.90E-6 = Kb = |QuinineH+|eq * |OH-|eq / |Quinine|eq
as the Acid-Base's Equilibrium of Quinine in its aqueoussolutions

ii) 1.0E-14 = Kw = |H+|eq * |OH-|eq
as the Self-Ionization's Equilibrium of Water

iii) 0.1800 = |Quinine|0 = |QuinineH+|eq + |Quinine|eq
as the Mass Balancement of Quinine

iv) |H+|eq + |QuinineH+|eq = |OH-|eq
as the Electrical Charge Balancement

These FOUR EQUATIONs form a Mathematical System leading toNumerical Solutions related to MATHEMATICAL UNKNOWNs, e.g. |H+|eq =1.5E-5 M or pH2 = 4.8. /

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