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Chapter 16: Problem 112

What is the \(\mathrm{pH}\) of a solution that is \(2.5 \times 10^{-9} \mathrm{M}\) in NaOH? Does your answer make sense?

Short Answer

Expert verified
The pH of a solution that is \(2.5 \times 10^{-9}M\) in NaOH can be found by first calculating the pOH (\(-\log_{10}(2.5 \times 10^{-9})\)), then subtracting it from 14 (pH + pOH = 14) to find the pH value. The calculated pH is approximately 8.6, which makes sense since basic solutions have pH values greater than 7.

Step by step solution

01

Write the ionization equation of NaOH

Since NaOH is a strong base, it ionizes completely in water to produce the hydroxide ions, OH-. The ionization equation of NaOH in water is: NaOH 鈫� Na鈦� + OH鈦� Given the concentration of NaOH is \(2.5 \times 10^{-9}M\), this means [OH鈦籡 is also \(2.5 \times 10^{-9}M\).
02

Calculate pOH

pOH is calculated using the equation: pOH = -log鈧佲個[OH鈦籡 Plug in the [OH鈦籡 value: pOH = -log鈧佲個(\(2.5 \times 10^{-9}\))
03

Calculate pH

Next, calculate the pH using the relationship between pH and pOH: pH + pOH = 14 Rearrange the equation to find pH: pH = 14 - pOH Plug in the pOH value calculated in step 2: pH 鈮� 14 - (-log鈧佲個(\(2.5 \times 10^{-9}\)))
04

Evaluate the answer for sense-check

Calculate the pH value: pH 鈮� 8.6 The pH value should be greater than 7 for basic solutions, so the pH value of 8.6 makes sense and confirms that the solution is basic.

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Key Concepts

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

Understanding Strong Bases
Strong bases are substances that completely dissociate into their ions in water. This full ionization means they break apart entirely, releasing hydroxide ions, denoted as OH鈦�. NaOH, commonly known as sodium hydroxide, is a classic example of a strong base.

If you add NaOH to water, it splits into Na鈦� and OH鈦� ions very readily. This total disassociation leads to high concentrations of hydroxide ions, which are pivotal in determining the basic nature of the solution. It's important to remember:
  • Strong bases like NaOH dissociate completely in water.
  • They result in a higher concentration of OH鈦� ions.
  • This complete ionization is key to their potent basic characteristics.
Understanding the nature of strong bases helps in predicting how a solution will behave in terms of its acid-base properties.
The Ionization Equation
The ionization equation of a strong base like NaOH is straightforward because it involves complete dissociation. Here鈥檚 how it looks:

NaOH 鈫� Na鈦� + OH鈦�

In this equation:
  • NaOH, when added to water, separates completely into Na鈦� (sodium ions) and OH鈦� (hydroxide ions).
  • The concentration of OH鈦� in the solution directly equals the initial concentration of NaOH, thanks to this complete ionization.
This equation holds substantial importance because it highlights the source of the solution's basicity. By breaking down the compound fully, the resulting hydroxide ions contribute significantly to the pH of the solution.
Steps to pOH Calculation
Calculating pOH is a key part of determining a solution's pH, especially when dealing with basic solutions. For bases, we frequently use the concentration of OH鈦� ions to find pOH. The formula to calculate pOH is:

pOH = -log鈧佲個[OH鈦籡

In this approach:
  • Take the concentration of OH鈦� ions, which is equal to the initial concentration of the base due to complete ionization.
  • Apply the logarithmic function: pOH = -log鈧佲個( 2.5 脳 10 ^{-9} ).
The resulting value represents the measure of hydroxide ions present, giving us a clearer understanding of the solution鈥檚 basic trait. The lower the pOH, the more basic the solution, allowing us to easily transition to finding the pH value with ease.
Determining Basic Solutions
Basic solutions are characterized by a pH greater than 7. This is derived from the pH scale, which spans from 0 to 14. For any basic solution:

pH + pOH = 14

When you know the pOH, you can determine the pH using the transformation: pH = 14 - pOH.

Let's explore:
  • After calculating the pOH, use the equation above to find the pH.
  • In our case, the result was pH 鈮� 8.6, giving a clear indication of a basic solution.
  • Basic solutions are fundamental in various chemical reactions and processes, and the pH scale helps in identifying the extent of their basicity.
Thus, understanding the pH of a basic solution, like one containing NaOH, is essential for many practical and theoretical chemistry applications.

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Most popular questions from this chapter

Arrange the following \(0.10 \mathrm{M}\) solutions in order of increasing acidity (decreasing \(\mathrm{pH}\) ): (i) \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) (ii) \(\mathrm{NaNO}_{3}\), (iii) \(\mathrm{CH}_{3} \mathrm{COONH}_{4}\), (iv) \(\mathrm{NaF}\), (v) \(\mathrm{CH}_{3} \mathrm{COONa}\)

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{pK}_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of this substance?

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 M ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;(\mathrm{c}) \mathrm{a}\) solution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l) .(\mathrm{c})\) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-} ?\) What is its conjugate base?

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-}\), (b) \(\mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-},(\mathrm{c}) \mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)
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