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Chapter 19: Problem 49

Calculate the pH during the titration of \(40.00 \mathrm{~mL}\) of \(0.1000 \mathrm{M}\) HCl with \(0.1000 M\) NaOH solution after each of the following additions of base: (a) \(0 \mathrm{~mL}\) (b) \(25.00 \mathrm{~mL}\) (c) \(39.00 \mathrm{~mL}\) (d) \(39.90 \mathrm{~mL}\) (e) \(40.00 \mathrm{~mL}\) (f) \(40.10 \mathrm{~mL}\) (g) \(50.00 \mathrm{~mL}\)

Short Answer

Expert verified
pH values are: (a) 1.00, (b) 1.64, (c) 2.89, (d) 3.90, (e) 7.00, (f) 10.10, (g) 12.04.

Step by step solution

01

Initial Setup

Identify the volume and molarity of HCl and NaOH. We have 40.00 mL of 0.1000 M HCl and we are adding 0.1000 M NaOH solution.
02

Calculation for each addition

For each given volume of NaOH added, calculate the pH accordingly.
03

Part (a): 0 mL NaOH added

Since no NaOH has been added, we only have 0.1000 M HCl: pH = -log[HCl] pH = -log(0.1000) pH = 1.00
04

Part (b): 25.00 mL NaOH added

Determine moles of HCl and NaOH: moles HCl = 0.0400 L * 0.1000 M = 0.0040 mol moles NaOH = 0.0250 L * 0.1000 M = 0.0025 mol Moles HCl remaining = 0.0040 mol - 0.0025 mol = 0.0015 mol New volume = 40.00 mL + 25.00 mL = 65.00 mL = 0.0650 L New concentration of HCl = 0.0015 mol / 0.0650 L = 0.0231 M pH = -log(0.0231) = 1.64
05

Part (c): 39.00 mL NaOH added

Determine moles of HCl and NaOH: moles HCl = 0.0400 L * 0.1000 M = 0.0040 mol moles NaOH = 0.0390 L * 0.1000 M = 0.0039 mol Moles HCl remaining = 0.0040 mol - 0.0039 mol = 0.0001 mol New volume = 40.00 mL + 39.00 mL = 79.00 mL = 0.0790 L New concentration of HCl = 0.0001 mol / 0.0790 L = 0.00127 M pH = -log(0.00127) = 2.89
06

Part (d): 39.90 mL NaOH added

Determine moles of HCl and NaOH: moles HCl = 0.0400 L * 0.1000 M = 0.0040 mol moles NaOH = 0.0399 L * 0.1000 M = 0.00399 mol Moles HCl remaining = 0.0040 mol - 0.00399 mol = 0.00001 mol New volume = 40.00 mL + 39.90 mL = 79.90 mL = 0.0799 L New concentration of HCl = 0.00001 mol / 0.0799 L = 1.25 x 10^{-4} M pH = -log(1.25 x 10^{-4}) = 3.90
07

Part (e): 40.00 mL NaOH added

Moles of HCl = Moles of NaOH Complete neutralization occurs pH = 7 (Neutral solution)
08

Part (f): 40.10 mL NaOH added

Determine moles of NaOH in excess: moles HCl = 0.0040 mol moles NaOH = 0.0401 L * 0.1000 M = 0.00401 mol Moles NaOH in excess = 0.00401 mol - 0.0040 mol = 0.00001 mol New volume = 40.00 mL + 40.10 mL = 80.10 mL = 0.0801 L New concentration of OH⁻ = 0.00001 mol / 0.0801 L = 1.25 x 10^{-4} M pOH = -log(1.25 x 10^{-4}) = 3.90 pH = 14 - pOH = 14 - 3.90 = 10.10
09

Part (g): 50.00 mL NaOH added

Determine moles of NaOH in excess: moles HCl = 0.0040 mol moles NaOH = 0.0500 L * 0.1000 M = 0.0050 mol Moles NaOH in excess = 0.0050 mol - 0.0040 mol = 0.0010 mol New volume = 40.00 mL + 50.00 mL = 90.00 mL = 0.0900 L New concentration of OH⁻ = 0.0010 mol / 0.0900 L = 0.0111 M pOH = -log(0.0111) = 1.96 pH = 14 - pOH = 14 - 1.96 = 12.04

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Key Concepts

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

acid-base titration
Acid-base titration is a common laboratory method used to determine the concentration of an acid or base solution by neutralizing it with a titrant of known concentration. The process involves slowly adding a base (like NaOH) to an acid (like HCl) while monitoring the pH changes. The primary goal is to reach the equivalence point, where the amount of titrant added exactly neutralizes the acid or base solution.
Neutralization is crucial because it allows for the precise calculation of unknown concentrations.
  • The titrant is added from a burette to the solution in the flask.
  • An appropriate indicator or pH meter is used for pH monitoring.
  • The equivalence point is identified when the pH undergoes a significant change, representing full neutralization.
pH calculation
Calculating the pH during an acid-base titration is fundamental to understanding the titration process. pH is a measure of the hydrogen ion concentration in a solution and is calculated using the formula:
\( pH = -\log[H^+] \). This formula helps determine the acidity or basicity of the solution.
  • Before any base is added, the pH is determined solely by the acid's concentration.
  • As the base is added, the acid is partially neutralized, altering the H\textsuperscript{+} concentration and therefore the pH.
  • Identifying the pH at various volumes of titrant added helps track progress toward the equivalence point.
For instance, in this titration, the pH calculations at different stages (like before any NaOH is added, halfway points, and slight excess points) provide a clear picture of how the solution's pH changes.
molarity
Molarity (M) is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. It is a key concept in titration because it helps quantify the amounts of reactants.
In the context of our exercise:
$$ \text {Molarity} = \frac {moles \text { of solute}} {volume \text {of solution in liters}}$$
  • Initial molarity of HCl is given as 0.1000 M in 40.00 mL (0.0400 L).
  • As NaOH is added, its molarity contributes to the overall neutralization process.
  • The change in molarity of HCl or excess NaOH post-neutralization is recalculated to determine the solution's pH.
Accurate molarity calculations are essential to determine the equivalence point and subsequent pH accurately.
neutralization reaction
A neutralization reaction is a chemical reaction where an acid reacts with a base to form water and a salt. In titration, this reaction is used to determine the point at which the amount of acid equals the amount of base (equivalence point).
The balanced chemical equation for a typical neutralization reaction is:
\[ HCl + NaOH \rightarrow NaCl + H_2O \]
  • HCl (hydrochloric acid) provides H\textsuperscript{+} ions.
  • NaOH (sodium hydroxide) provides OH\textsuperscript{−} ions.
  • When H\textsuperscript{+} and OH\textsuperscript{−} combine, they form water (H\textsubscript{2}O).
  • The remaining Na\textsuperscript{+} and Cl\textsuperscript{−} ions form the salt NaCl.
This reaction is crucial for determining the titration's progress. Each addition of the base neutralizes part of the acid until complete neutralization is achieved, marked by the steep pH change.
stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. In titration, it’s crucial for accurately determining the equivalence point and final pH. In our problem:
  • Initially, calculate moles of HCl using its molarity and volume.
  • Do the same for NaOH as it is added in known volumes.
  • Subtract moles of NaOH from moles of HCl to find remaining moles of HCl or excess NaOH.
For example:
When 25.00 mL of 0.1000 M NaOH is added to 40.00 mL of 0.1000 M HCl:
\( \text{moles of HCl} = 0.0400L \times 0.1000M = 0.0040 \text{mol} \)
\( \text{moles of NaOH} = 0.0250L \times 0.1000M = 0.0025 \text{mol} \)
\( \text{remaining moles of HCl} = 0.0040 \text{mol} - 0.0025 \text{mol} = 0.0015 \text{mol} \)
This step-by-step approach facilitates accurate pH calculations after each titrant addition, guiding you through to the equivalence point and beyond.

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Most popular questions from this chapter

Find the molar solubility of \(\mathrm{BaCrO}_{4}\left(K_{\mathrm{sp}}=2.1 \times 10^{-10}\right) \mathrm{in}\) (a) pure water and (b) \(1.5 \times 10^{-3} M \mathrm{Na}_{2} \mathrm{CrO}_{4}\)

What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a propanoic acidpropanoate buffer that consists of \(0.35 M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\) and \(0.15 M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}\right.\) of propanoic acid \(\left.=1.3 \times 10^{-5}\right) ?\)

What is the component concentration ratio, \(\left[\mathrm{BrO}^{-}\right] /[\mathrm{HBrO}]\), of a buffer that has a \(\mathrm{pH}\) of \(7.95\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HBrO}=2.3 \times 10^{-9}\right) ?\)

What is the purpose of an acid-base buffer?

Gout is caused by an error in metabolism that leads to a buildup of uric acid in body fluids, which is deposited as slightly soluble sodium urate \(\left(\mathrm{C}_{5} \mathrm{H}_{3} \mathrm{~N}_{4} \mathrm{O}_{3} \mathrm{Na}\right)\) in the joints. If the extracellular \(\left[\mathrm{Na}^{+}\right]\) is \(0.15 \mathrm{M}\) and the solubility of sodium urate is \(0.085 \mathrm{~g} / 100 . \mathrm{mL}\) what is the minimum urate ion concentration (abbreviated [Ur-]) that will cause a deposit of sodium urate?
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