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Chapter 17: Problem 46

Determine the following characteristics of the titration curve for \(20.0 \mathrm{mL}\) of \(0.275 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq})\) titrated with \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) (a) the initial \(\mathrm{pH}\) (b) the volume of \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) at the equivalence point (c) the \(\mathrm{pH}\) at the half-neutralization point (d) the \(\mathrm{pH}\) at the equivalence point

Short Answer

Expert verified
The initial pH of the solution is around 11.23. The volume of 0.325 M HI(aq) required at the equivalence point is approximately 16.92 mL. The pH at the half-neutralization point is around 8.75 and the pH at the equivalence point is approximately 5.27.

Step by step solution

01

Calculate Initial pH

The initial pH is calculated using the concentration of the weak base; NH3 is a weak base and thus will form a basic solution when dissolved in water. The expression for the ionization of NH3 in water is given as: NH3 +H2O ↔ NH4+ + OH-. From this, the Kb expression can be formed as \([NH4^+][OH^-]/[NH3]\). Replace known values into Kb expression and find [OH-]. Next, Use the OH- concentration to calculate pOH using the formula -log[OH-] and then find pH by subtracting pOH from 14.
02

Calculate the volume of the titrant at equivalence point

The volume of the strong acid needed to completely neutralize weak base is calculated by applying the equivalence of moles at the equivalence point. According to the titration stoichiometry, 1 mole of HI neutralizes 1 mole of NH3. From this, n(NH3) = V(NH3)*C(NH3) and n(HI) = V(HI)*C(HI). Because at the equivalence point n(HI)=n(NH3), we can express V(HI) as n(NH3)/C(HI). Plug in the values and calculate V(HI).
03

Calculate pH at the half neutralization point

At the half-neutralization point, the pOH is equal to the pKb of the weak base. Hence, we can calculate the pKb of NH3 using the given Kb value in textbooks or chemical literature, then calculate pH, pH = 14 - pKb.
04

Calculate pH at the equivalence point

At the equivalence point, all the weak base has been reacted with the strong acid to produce its conjugate acid (NH4+). This situation represents a solution of a weak acid, so we have to determine pH of this solution. The expression for the ionization of NH4+ in water is given as: NH4+ ↔ NH3 + H3O+. From this, the Ka expression can be formed as [NH3][H3O+]/[NH4+]. Replace the known values into Ka expression and find [H3O+]. Convert H3O+ concentration to pH by using -log[H3O+].

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Most popular questions from this chapter

Calculate the pH at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.160 \mathrm{M} \mathrm{HCl}\) when (a) \(10.00 \mathrm{mL}\) and \((\mathrm{b}) 15.00 \mathrm{mL}\) of 0.242 M KOH have been added.

Indicate whether you would expect the equivalence point of each of the following titrations to be below, above, or at \(\mathrm{pH}\) 7. Explain your reasoning. (a) \(\mathrm{NaHCO}_{3}(\mathrm{aq})\) is titrated with \(\mathrm{NaOH}(\mathrm{aq})\) (a) (b) \(\mathrm{HCl}(\mathrm{aq})\) is titrated with \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{KOH}(\mathrm{aq})\) is titrated with HI(aq).

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\text { aq) requires } 31.15 \mathrm{mL}\) of \(0.2420 \mathrm{M}\) KOH for titration to the second equivalence point. What is the molarity of the \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) ?\)

What is the pH of a solution prepared by dissolving \(8.50 \mathrm{g}\) of aniline hydrochloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\right)\) in \(750 \mathrm{mL}\) of \(0.215 \mathrm{M}\) aniline, \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right) ?\) Would this solution be an effective buffer? Explain.

A very common buffer agent used in the study of biochemical processes is the weak base TRIS, \(\left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{2},\) which has a \(\mathrm{pK}_{\mathrm{b}}\) of 5.91 at \(25^{\circ} \mathrm{C} . \mathrm{A}\) student is given a sample of the hydrochloride of TRIS together with standard solutions of \(10 \mathrm{M}\) NaOH and HCl. (a) Using TRIS, how might the student prepare 1 L of a buffer of \(\mathrm{pH}=7.79 ?\) (b) In one experiment, 30 mmol of protons are released into \(500 \mathrm{mL}\) of the buffer prepared in part (a). Is the capacity of the buffer sufficient? What is the resulting pH? (c) Another student accidentally adds \(20 \mathrm{mL}\) of \(10 \mathrm{M}\) HCl to 500 mL of the buffer solution prepared in part (a). Is the buffer ruined? If so, how could the buffer be regenerated?
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