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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 H鈧傾 and OH-

To find the pH, we first need to calculate the moles of H鈧傾 and OH- in the solution at 125 mL of KOH added. We have 50.0 mL of 0.10 M H鈧傾 and 125 mL of 0.10 M KOH. Moles of H鈧傾 = 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 H鈧傾 to form water and the conjugate base. Since we have added more moles of OH- than H鈧傾, the solution will have excess OH-, meaning it will be basic. H鈧傾 + OH- 鈫 H鈧侫- + H鈧侽
03

Calculate moles of remaining species

Since all the moles of H鈧傾 (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 H鈧傾 + 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鈦 concentration = 10^(-pH) = 10^(-3.3) = 5.01 脳 10鈦烩伌 M
06

Determine reactions at the equivalence point

At the equivalence point, all moles of H鈧傾 are converted to H鈧侫-, so the solution will be at its second acidic dissociation point. We will use the second dissociation constant, Ka鈧. H鈧侫- + H鈧侽 鈫 HA虏鈦 + H鈧僌鈦 Ka鈧 = [HA虏鈦籡[H鈧僌鈦篯/[H鈧侫-] Using Ka鈧 = 1 脳 10鈦烩伕 and substituting [H鈧僌鈦篯 = 5.01 脳 10鈦烩伌 M, we can find the ratio of moles of H鈧侫炉 to moles of HA虏鈦 at pH = 3.30.
07

Calculate the volume of KOH needed

Since we have found the ratio of H鈧侫炉 to HA虏鈦 at pH = 3.30, we can calculate the volume of KOH needed to reach this pH. 1 脳 10鈦烩伕 = (5.01 脳 10鈦烩伌)^2 / x x = [HA虏鈦籡/[H鈧侫炉] = 1 脳 10^4 [H鈧傾] = 0.10 M, and at pH = 3.30, [H鈧傾] = 0.10 - x 0.10 - x = x(1 脳 10^4) x = [H鈧傾] = 9.09 脳 10鈦烩伒 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 [H鈧傾]/[KOH] V_KOH = (9.09 脳 10鈦烩伒)/(0.10) = 9.09 脳 10鈦烩伌 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 H鈧傾 and OH-

First, find the moles of H鈧傾 and OH- in the solution: Moles of H鈧傾 = 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 H鈧傾 > moles of OH-, the solution is acidic.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH calculation
The concept of pH is crucial in chemistry, as it measures the acidity or basicity of a solution. The pH scale ranges from 0 to 14:
  • A pH less than 7 means the solution is acidic.
  • A pH of 7 indicates a neutral solution.
  • A pH greater than 7 means the solution is basic.
To calculate the pH from the concentration of hydrogen ions (H^+), you can use the formula:\[pH = -\log[H^+]\]For example, if you know the concentration of H^+ is 5 \times 10^{-4} \text{ M}, the pH is calculated as follows:\[pH = -\log(5 \times 10^{-4}) \approx 3.30\]In a basic solution, like the one formed in the KOH and H_3A titration, instead of calculating pH directly, you can find \text{pOH} first, and then derive the pH using:\[pH = 14 - \text{pOH}\]Understanding these calculations is essential for analyzing the acidity or basicity of solutions during titrations.
neutralization reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt. This type of chemical reaction generally involves the following equation:\[\mathrm{HA} + \mathrm{BOH} \rightarrow \mathrm{BA} + \mathrm{H_2O}\]In the exercise provided, H_3A, a weak acid, reacts with KOH, a strong base.
  • As KOH is added to H_3A, it reacts to neutralize the acid, forming water and the conjugate base (H_2A^-).
  • This process continues until all the H_3A is converted, or excess KOH is added, creating a basic solution due to leftover OH^- ions.
In titration processes, the point at which the amount of acid equals the amount of base is called the equivalence point. In acid-base titrations, finding this point is crucial, as it helps to determine concentrations and the PH of the solution.
  • At equivalence, assuming complete reaction, the moles of H^+ from the acid equals the moles of OH^- from the base.
  • In the problem, determining excess OH^- means any additional base will increase the pH, indicating a basic solution.
weak acids
Weak acids are characterized by their partial dissociation in water, which means they do not release all their hydrogen ions into the solution.
  • This property contrasts sharply with strong acids, which completely dissociate in water.
  • The strength of a weak acid is represented by its acid dissociation constant (K_a), which quantifies its tendency to donate H^+ ions in the solution.
In the case of H_3A, a triprotic acid, it dissociates in steps, having three distinct K_a values corresponding to the release of each hydrogen ion:
  • \(K_{\text{a}_1} = 5.0 \times 10^{-4}\) for the first hydrogen.
  • \(K_{\text{a}_2} = 1.0 \times 10^{-8}\) for the second hydrogen.
  • \(K_{\text{a}_3} = 1.0 \times 10^{-11}\) for the third hydrogen.
The gradual nature of dissociation in weak acids impacts the titration curve and necessitates careful analysis to determine pH changes. Understanding the behavior of weak acids during titration helps in assessing their buffering capacity and predicting the pH of solutions during different stages of reactions.

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Most popular questions from this chapter

Sketch the titration curves for a diprotic acid titrated by a strong base and a triprotic acid titrated by a strong base. List the major species present at various points in each curve. In each curve, label the halfway points to equivalence. How do you calculate the pH at these halfway points?

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125\(M \mathrm{NaOH}\) , requiring 16.00 \(\mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding 2.00 \(\mathrm{mL}\) . NaOH was 6.912 . Calculate \(K_{\mathrm{a}}\) for the weak acid.

Calculate the ph of each of the following solutions. $$ \begin{array}{l}{\text { a. } 0.100 M \text { propanoic acid }\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)} \\ {\text { b. } 0.100 M \text { sodium propanoate }\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)} \\ {\text { c. pure } \mathrm{H}_{2} \mathrm{O}}\end{array} $$ $$ \begin{array}{l}{\text { d. a mixture containing } 0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} \text { and } 0.100 \mathrm{M}} \\\ {\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}}\end{array} $$

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 \(\mathrm{L}\) of each of the following solutions: a. 0.050\(M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)\) and 0.080\(M\) sodium propanoate b. 0.50\(M\) propanoic acid and 0.80\(M\) sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NaOH}\) . The student then adds 13.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) . The pH of the resulting solution is \(4.70 .\) How is the value of \(\mathrm{p} K_{2}\) for the unknown acid related to 4.70\(?\)

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