Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Both sodium hydrogen carbonate (sodium bicarbonate) and sodium hydroxide can be used to neutralize acid spills. What is the pH of \(1.00 \mathrm{M} \mathrm{NaHCO}_{3}(\mathrm{aq})\) and of \(1.00 \mathrm{M} \mathrm{NaOH}(\mathrm{aq}) ?\) On a per-liter basis, do these two solutions have an equal capacity to neutralize acids? Explain. On a per-gram basis, do the two solids, \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{NaOH}(\mathrm{s}),\) have an equal capacity to neutralize acids? Explain. Why do you suppose that \(\mathrm{NaHCO}_{3}\) is often preferred to \(\mathrm{NaOH}\) in neutralizing acid spills?

Short Answer

Expert verified
The pH of a 1.00 M \(\mathrm{NaHCO}_{3}\) solution is around 8.3 and for a 1.00 M \(\mathrm{NaOH}\) solution is 14. On a per-liter basis, \(\mathrm{NaOH}\) has a greater capacity to neutralize acids than \(\mathrm{NaHCO}_{3}\) because it dissociates completely in the solution. On a per-gram basis, \(\mathrm{NaOH}\) also has a greater capacity to neutralize acids because it has a lower molar mass and thus provides more OH- ions per gram. \(\mathrm{NaHCO}_{3}\) is often preferred as it is safer to handle, less likely to result in injury, and provides an indicator (CO2 gas) when reacting with acids.

Step by step solution

01

Calculate pH of Sodium Hydroxide (\(\mathrm{NaOH}\)) solution

Using the formula: pH = -log[H+], and knowing that a 1.00 M \(\mathrm{NaOH}\) solution is a strong base and it will completely dissociate into \(\mathrm{Na}^{+}\) and \(\mathrm{OH}^{-}\) ions resulting in [OH-] = 1.00 M. The relationship between [H+] and [OH-] is represented as Kw=[H+][OH-], where Kw is the ion product of water, equal to \(1 * 10^{-14}\). Solving for [H+], we get [H+] = Kw/[OH-] = \(1 * 10^{-14}/1.00\). Using this value to solve for pH, we get pH = -log[H+] = 14.
02

Calculate pH of Sodium Bicarbonate (\(\mathrm{NaHCO}_{3}\)) solution

Considering \(\mathrm{NaHCO}_{3}\) as a weak base, and knowing that it partially dissociates in water to form bicarbonate ions \(HCO_{3}^{-}\) and hydronium ions \(H_{3}O^{+}\). The reaction can be written as: \(\mathrm{NaHCO}_{3}\) + H2O <-> \(H_{3}O^{+}\) + \(\mathrm{HCO}_{3}^{-}\). Now the equation for the bicarbonate ion acting as an acid can be written as: \(\mathrm{HCO}_{3}^{-}\) <-> \(H_{3}O^{+}\) + \(\mathrm{CO}_{3}^{2-}\). By setting up an equilibrium constant expression for this reaction with the known \(Ka\) and \(Kb\) values for the bicarbonate ion, the [H+] concentration can be solved. Using this \(H^{+}\) concentration, the pH of the solution can be calculated, which results to be approximately 8.3.
03

Neutralizing Capacity on a Per-Litre Basis

To neutralize an acid, a base must provide OH- ions. A 1.00 M solution of NaOH will provide a greater number of OH- ions (and hence a greater acid-neutralizing capacity) than a 1.00 M solution of NaHCO3, because the NaOH dissociates completely while NaHCO3 only partially dissociates to provide OH- ions.
04

Neutralizing Capacity on a Per-gram Basis

To neutralize an acid, a base must provide OH- ions. Comparing the molar masses (NaHCO3: 84 g/mol, NaOH: 40 g/mol), we find that per gram, NaOH provides a greater number of OH- ions (and hence a greater acid-neutralizing capacity) than NaHCO3.
05

Preference of \(\mathrm{NaHCO}_{3}\) over \(\mathrm{NaOH}\) for Acid Spills

\(\mathrm{NaHCO}_{3}\) is often preferred over \(\mathrm{NaOH}\) in neutralizing acid spills because it is a weaker base as compared to \(\mathrm{NaOH}\), so it is inherently safer to handle and less likely to cause injury due to excessive neutralization (over-shooting the pH to highly basic). It also releases carbon dioxide gas when it reacts with acids, providing an indicator of reaction progress.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding pH Calculation
Calculating the pH of a solution is essential to understanding its acidity or basicity. The pH scale ranges from 0 to 14. A pH less than 7 indicates acidity, while a pH greater than 7 indicates basicity.

For strong bases like sodium hydroxide (NaOH), the pH calculation involves complete dissociation into ions in water. NaOH dissociates into sodium (Na+) and hydroxide (OH-) ions. In a 1.00 M NaOH solution, the concentration of OH- ions is 1.00 M.

The relationship between the concentrations of hydrogen ions \[ \text{[H+]} \] and hydroxide ions \[ \text{[OH-]} \] is given by the water dissociation constant \[ K_w = 10^{-14} \]. Solving for [H+], we use \[ \text{[H+]} = \frac{K_w}{\text{[OH-]}} \].

For NaOH: \[ \text{[H+]} = \frac{10^{-14}}{1.00} = 10^{-14} \]. The pH is calculated as \[ pH = -\log(\text{[H+]}) = 14 \].
  • For weak bases like sodium bicarbonate (NaHCO3), only partial dissociation occurs.
  • This results in a basic but less high pH value, around 8.3 for a 1.00 M solution.
The Role of Sodium Hydroxide
Sodium hydroxide (NaOH) is a powerful, strong base widely used in chemical processes. It fully dissociates in water, releasing hydroxide ions \[ \text{OH-} \].

It is highly effective in neutralizing acids due to this complete dissociation, making it a go-to choice in many industrial applications. However, handling NaOH requires caution. Its strong basic nature can cause chemical burns and should be managed with appropriate safety measures.
  • Effective in neutralizing strong acids.
  • Provides a high concentration of hydroxide ions.
Despite its efficiency, it is not always preferred for neutralizing acid spills because it can cause the pH to rise too rapidly, leading to a strongly alkaline, rather than neutral, solution.
Understanding Sodium Bicarbonate
Sodium bicarbonate (NaHCO3), often known as baking soda, is a versatile, weak base. In water, it partially dissociates to produce bicarbonate \[ \text{HCO-3} \] and minimal hydroxide ions.

This mild dissociation results in a moderate pH, making NaHCO3 suitable for gentle pH adjustments. Its benefits include:
  • Being safer to handle than stronger bases.
  • Less risk of drastically overshooting the target pH.
  • Production of carbon dioxide gas when reacting with acids, providing a visual cue of reaction efficacy.
This makes sodium bicarbonate a safer option for neutralizing acid spills, where a more controlled reaction is desired to avoid creating a highly basic environment.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 Lof a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is to (a) raise the pH very slightly; (b) lower the pH very slightly; (c) raise the pH by several units; (d) lower the pH by several units.

Because an acid-base indicator is a weak acid, it can be titrated with a strong base. Suppose you titrate \(25.00 \mathrm{mL}\) of a \(0.0100 \mathrm{M}\) solution of the indicator \(p\) -nitrophenol, \(\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{NO}_{2},\) with \(0.0200 \mathrm{M} \mathrm{NaOH}\) The \(\mathrm{p} K_{\mathrm{a}}\) of \(p\) -nitrophenol is \(7.15,\) and it changes from colorless to yellow in the pH range from 5.6 to 7.6 (a) Sketch the titration curve for this titration. (b) Show the pH range over which \(p\) -nitrophenol changes color. (c) Explain why \(p\) -nitrophenol cannot serve as its own indicator in this titration.

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a solution that is (a) \(0.0062 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.0105 \mathrm{M} \mathrm{BaCl}_{2} ;\) (b) \(0.315 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) and \(0.486 \mathrm{M} \mathrm{NH}_{3} ;\) (c) \(0.196 \mathrm{M} \mathrm{NaOH}\) and \(0.264 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}\)

Indicate which of the following aqueous solutions are buffer solutions, and explain your reasoning. (a) \(0.100 \mathrm{M} \mathrm{NaCl}\) (b) \(0.100 \mathrm{M} \mathrm{NaCl}-0.100 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (c) \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}-0.150 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\) (d) \(0.100 \mathrm{M} \mathrm{HCl}-0.050 \mathrm{M} \mathrm{NaNO}_{2}\) (e) \(0.100 \mathrm{M} \mathrm{HCl}-0.200 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\) (f) \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}-0.125 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}\)

Using appropriate equilibrium constants but without doing detailed calculations, determine whether a solution can be simultaneously: (a) \(0.10 \mathrm{M} \mathrm{NH}_{3}\) and \(0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl},\) with \(\mathrm{pH}=6.07\) (b) \(0.10 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(0.058 \mathrm{M} \mathrm{HI}\) (c) \(0.10 \mathrm{M} \mathrm{KNO}_{2}\) and \(0.25 \mathrm{M} \mathrm{KNO}_{3}\) (d) \(0.050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.65 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (e) \(0.018 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) and \(0.018 \mathrm{M} \mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}\) with \(\mathrm{pH}=4.20\) (f) \(0.68 \mathrm{M} \mathrm{KCl}, 0.42 \mathrm{M} \mathrm{KNO}_{3}, 1.2 \mathrm{M} \mathrm{NaCl},\) and \(0.55 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) with \(\mathrm{pH}=6.4\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free