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Chapter 15: Problem 88

Consider the titration of 50.0 \(\mathrm{mL}\) of 0.10$M \mathrm{H}_{3} \mathrm{A}\left(K_{\mathrm{a}},=\right.\( \)5.0 \times 10^{-4}, K_{\mathrm{a}_{2}}=1.0 \times 10^{-8}, K_{\mathrm{a}_{2}}=1.0 \times 10^{-11}$ ) titrated by 0.10\(M \mathrm{KOH}\) a. Calculate the pH of the resulting solution at 125 \(\mathrm{mL}\) of KOH added. b. At what volume of KOH added does pH \(=3.30 ?\) c. At 75.0 \(\mathrm{mL}\) of KOH added, is the solution acidic or basic?

Short Answer

Expert verified
The pH of the resulting solution at 125 mL of KOH added is approximately 12.63. To reach a pH of 3.30, approximately 0.909 mL of KOH must be added. At 75.0 mL of KOH added, the solution is acidic.

Step by step solution

01

Calculate moles of H₃A and OH-

To find the pH, we first need to calculate the moles of H₃A and OH- in the solution at 125 mL of KOH added. We have 50.0 mL of 0.10 M H₃A and 125 mL of 0.10 M KOH. Moles of H₃A = 0.10 M * 50.0 mL * (1 L/1000 mL) = 0.005 mol Moles of OH- = 0.10 M * 125 mL * (1 L/1000 mL) = 0.0125 mol
02

Determine the neutralization reaction

The strong base KOH will react with the weak acid H₃A to form water and the conjugate base. Since we have added more moles of OH- than H₃A, the solution will have excess OH-, meaning it will be basic. H₃A + OH- → H₂A- + H₂O
03

Calculate moles of remaining species

Since all the moles of H₃A (0.005 mol) will react with 0.005 mol of OH-, we are left with 0.0075 mol of OH- (0.0125 - 0.005). The solution is basic, so we can find the concentration of OH- in the solution. Total volume = initial volume of H₃A + volume of KOH = 50 mL + 125 mL = 175 mL = 0.175 L Concentration of OH- = (0.0075 mol)/0.175 L = 0.04286 M
04

Calculate the pH

Since the solution is basic, we can use the concentration of OH- to determine the pOH and finally the pH. pOH = -log(0.04286) = 1.367 pH = 14 - pOH = 14 - 1.367 = 12.633 The pH of the resulting solution at 125 mL of KOH added is approximately 12.63. b. At what volume of KOH added does pH = 3.30?
05

Calculate the hydrogen ion concentration

To find the volume of KOH added when the pH is 3.30, we first need to determine the hydrogen ion concentration: pH = 3.30 H⁺ concentration = 10^(-pH) = 10^(-3.3) = 5.01 × 10⁻⁴ M
06

Determine reactions at the equivalence point

At the equivalence point, all moles of H₃A are converted to H₂A-, so the solution will be at its second acidic dissociation point. We will use the second dissociation constant, Ka₂. H₂A- + H₂O → HA²⁻ + H₃O⁺ Ka₂ = [HA²⁻][H₃O⁺]/[H₂A-] Using Ka₂ = 1 × 10⁻⁸ and substituting [H₃O⁺] = 5.01 × 10⁻⁴ M, we can find the ratio of moles of H₂A¯ to moles of HA²⁻ at pH = 3.30.
07

Calculate the volume of KOH needed

Since we have found the ratio of H₂A¯ to HA²⁻ at pH = 3.30, we can calculate the volume of KOH needed to reach this pH. 1 × 10⁻⁸ = (5.01 × 10⁻⁴)^2 / x x = [HA²⁻]/[H₂A¯] = 1 × 10^4 [H₃A] = 0.10 M, and at pH = 3.30, [H₃A] = 0.10 - x 0.10 - x = x(1 × 10^4) x = [H₃A] = 9.09 × 10⁻⁵ moles (at pH = 3.30) Since we are given the concentration of KOH, we can find the volume of KOH needed to reach pH = 3.30. V_KOH (in L) = moles [H₃A]/[KOH] V_KOH = (9.09 × 10⁻⁵)/(0.10) = 9.09 × 10⁻⁴ L, which is equivalent to 0.909 mL Therefore, the volume of KOH added to reach pH = 3.30 is approximately 0.909 mL. c. At 75.0 mL of KOH added, is the solution acidic or basic?
08

Determine the moles of H₃A and OH-

First, find the moles of H₃A and OH- in the solution: Moles of H₃A = 0.005 mol (as previously calculated for 50 mL of 0.10 M) Moles of OH- = 0.10 M * 75.0 mL * (1 L/1000 mL) = 0.0075 mol Since moles of H₃A > moles of OH-, the solution is acidic.

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Most popular questions from this chapter

A certain acetic acid solution has \(\mathrm{pH}=2.68\) . Calculate the volume of 0.0975 \(\mathrm{M} \mathrm{KOH}\) required to reach the equivalence point in the titration of 25.0 \(\mathrm{mL}\) of the acetic acid solution.

Some \(\mathrm{K}_{2} \mathrm{SO}_{3}\) and \(\mathrm{KHSO}_{3}\) are dissolved in 250.0 \(\mathrm{mL}\) of solution and the resulting \(\mathrm{pH}\) is \(7.25 .\) Which is greater in this buffer solution, the concentration of \(\mathrm{SO}_{3}^{2-}\) or the concentration of \(\mathrm{HSO}_{3}-7\) If \(\left[\mathrm{SO}_{3}^{2-}\right]=1.0 \mathrm{M}\) in this solution, calculate the concentration of \(\mathrm{HSO}_{3}\) .

Which of the following can be classified as buffer solutions? $$ \begin{array}{l}{\text { a. } 0.25 M \mathrm{HBr}+0.25 \mathrm{M} \mathrm{HOBr}} \\ {\text { b. } 0.15 \mathrm{M} \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}} \\ {\text { c. } 0.50 \mathrm{M} \mathrm{HOCl}+0.35 \mathrm{MKOCl}}\end{array} $$ $$ \begin{array}{l}{\text { d. } 0.70 M \mathrm{KOH}+0.70 \mathrm{M} \text { HONH_ }} \\ {\text { e. } 0.85 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}}\end{array} $$

A student dissolves 0.0100 mol of an unknown weak base in 100.0 \(\mathrm{mL}\) water and titrates the solution with 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) After 40.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

Tris (hydroxymethyl)aminomethane, commonly called TRIS or Trizma, is often used as a buffer in biochemical studies. Its buffering range is $\mathrm{pH} 7\( to \)9,\( and \)K_{\mathrm{b}}\( is \)1.19 \times 10^{-6}$ for the aqueous reaction $$ \left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{3}^{+}+\mathrm{OH}^{-} $$ a. What is the optimal pH for TRIS buffers? b. Calculate the ratio \([T R I S] /\left[T R I S H^{+}\right]\) at \(p H=7.00\) and at p H=9.00 c. A buffer is prepared by diluting 50.0 \(\mathrm{g}\) TRIS base and 65.0 \(\mathrm{g}\) TRIS hydrochloride (written as TRISHCl) to a total volume of 2.0 \(\mathrm{L}\) What is the pH of this buffer? What is the pH after 0.50 \(\mathrm{mL}\) of 12 \(\mathrm{MHCl}\) is added to a 200.0 -mL portion of the buffer?
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