Water in equilibrium with air contains \(4.4 \times 10^{-5} \% \mathrm{CO}_{2}\). The resulting carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}\), gives the solution a hydronium ion concentration of \(2.0\) \(\times 10^{-6} \mathrm{M}\), about 20 times greater than that of pure water. What is the \(\mathrm{pH}\) of the solution at \(298 \mathrm{~K} ?(\log 4.4=0.64\) \(\log 2=0.3\) ) (a) \(5.36\) (b) \(5.70\) (c) \(8.30\) (d) \(5.64\)

Short Answer

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5.70

Step by step solution

01

Understanding the Concept of pH

The pH of a solution is defined as the negative logarithm (base 10) of the hydronium ion concentration. The formula is given by: \[ \text{pH} = -\log{[H_3O^+]} \]
02

Calculating the pH

Given the hydronium ion concentration of \(2.0 \times 10^{-6} \mathrm{M}\), we use the pH formula to calculate the pH of the solution: \[ \text{pH} = -\log{[2.0 \times 10^{-6}]} \]
03

Using Logarithmic Properties

We can use the property that \(\log{(a \times b)} = \log{a} + \log{b}\). Let's break down the logarithm into two terms and use the given logarithmic values: \[ \text{pH} = -(\log{2} + \log{10^{-6}}) \] Substitute the given logarithmic values: \[ \text{pH} = - (0.3 + (-6)) \]
04

Compute the Final pH Value

Simplify the equation to find the pH value: \[ \text{pH} = 6 - 0.3 = 5.7 \]

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Key Concepts

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

Understanding the pH of a Solution
The concept of pH is a critical aspect of chemistry, especially when studying solutions and their acidic or basic nature. The pH is a scale used to quantify the acidity or basicity of a solution. It's essential to know that the pH scale ranges from 0 to 14 where a pH less than 7 indicates an acidic solution, pH equal to 7 indicates a neutral solution, and pH greater than 7 indicates a basic solution. The term 'pH' stands for 'potential of hydrogen' or 'power of hydrogen' and refers to the amount of hydrogen ions present in a solution. The formal definition of pH is the negative logarithm (to the base 10) of the hydronium ion concentration
(\( pH = -\log[H_3O^+] \)).
Understanding pH is fundamental because it affects various chemical reactions, biological processes, and even daily products like food and cleaning agents. Moreover, in environmental science, the pH of water bodies is a key indicator of their health, where extremely high or low pH values can lead to negative effects on aquatic life.
Interpreting Hydronium Ion Concentration
Hydronium ions (\( H_3O^+ \)) are positively charged ions that form when water molecules combine with hydrogen ions. The concentration of these hydronium ions in a solution is a direct measure of the solution's acidity. In the context of the given exercise, the hydronium ion concentration was found to be \(2.0 \times 10^{-6} \mathrm{M}\), which is relatively higher compared to pure water. This increased concentration is the result of dissolved carbon dioxide from air forming carbonic acid in water, which subsequently dissociates to produce these hydronium ions. Understanding this concentration is crucial as it is the starting point in calculating the pH of the solution and can give insights into the chemical properties and behaviour of the solution. For example, knowing the hydronium ion concentration helps in predicting the corrosive nature of the solution or how it would interact with other substances.
Applying Logarithmic Properties
Logarithmic properties are mathematical tools that help simplify calculations involving powers of 10, as often seen in pH calculations. One such property used in pH computation is the product rule for logarithms, which states that the log of a product is the sum of the logs (
\( \log(ab) = \log(a) + \log(b) \)).
For instance, in the exercise provided, the hydronium ion concentration given is in the form of a number times a power of ten. Using this property, we separate the logarithm of the number from the logarithm of the power of ten. Additionally, the fact that \( \log(10^n) = n \) is used to simplify the term involving \(10^{-6}\). Mastering these properties not only makes it possible to quickly compute pH without a calculator but also strengthens one’s overall algebraic skills, which are invaluable in the broader scope of scientific calculation.
Exploring Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time. In the context of the given problem, chemical equilibrium is reached when carbon dioxide dissolves in water, forming carbonic acid and eventually producing hydronium ions. The concentration of these ions influences the equilibrium constants for the dissolution of carbon dioxide and the dissociation of carbonic acid.
Understanding chemical equilibrium is vital to engineers and environmental scientists; for instance, when treating wastewater or assessing the carbon cycle in nature. By applying the principles of equilibrium, scientists can predict how changes in conditions like temperature or pressure will affect the direction and extent of chemical reactions.

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Most popular questions from this chapter

Ascorbic acid (vitamin \(\mathrm{C}\) ) is a diprotic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\). What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution? The acid ionization constants are \(K_{\mathrm{al}}=9.0 \times 10^{-5}\) and \(K_{\mathrm{a} 2}=1.6 \times 10^{-12} \cdot(\log 2=0.3, \log 3=0.48)\) (a) \(3.52\) (b) \(2.52\) (c) \(1.52\) (d) \(2.48\)

An amount of \(0.01\) moles of solid \(\mathrm{AgCN}\) is rendered soluble in 11 by adding just sufficient excess cyanide ion to form \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) and the concentration of free cyanide ion is \(2.5 \times 10^{-7} \mathrm{M}\). Determine \(\left[\mathrm{Ag}^{+}\right]\) in the solution neglecting hydrolysis of cyanide ion. \(K_{\text {diss }}\) for \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) \(=1.0 \times 10^{-20}\) (a) \(6.25 \times 10^{-9} \mathrm{M}\) (b) \(1.6 \times 10^{-9} \mathrm{M}\) (c) \(1.6 \times 10^{-7} \mathrm{M}\) (d) \(6.25 \times 10^{-7} \mathrm{M}\)

A solution contains \(0.1 \mathrm{M}-\mathrm{Mg}^{2+}\) and \(\begin{array}{lll}0.1 & \mathrm{M} & -\mathrm{Sr}^{2+} & \text { The concentration of }\end{array}\) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in solution is adjusted to \(0.05 \mathrm{M}\). Determine the \(\mathrm{pH}\) range which would permit the precipitation of \(\mathrm{SrCO}_{3}\) without any precipitation of \(\mathrm{MgCO}_{3} . \mathrm{H}^{+}\) ion concentration is controlled by external factors. Given: \(K_{\mathrm{sp}}\left(\mathrm{MgCO}_{3}\right)=4 \times 10^{-8} \mathrm{M}^{2}\); \(K_{\mathrm{sp}}\left(\mathrm{SrCO}_{3}\right)=9 \times 10^{-10} \mathrm{M}^{2} ; K_{\mathrm{a}, \text { overall }}\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) \(=5 \times 10^{-17} ; \log 2=0.3 ; \log 3=0.48 .\) (a) \(4.78\) to \(5.6\) (b) \(4.6\) to \(5.78\) (c) \(5.78\) to \(6.4\) (d) \(5.22\) to \(5.4\)

How much water must added to \(300 \mathrm{ml}\) of \(0.2 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) for the degree of dissociation of the acid to double? \(K_{\mathrm{a}}\) for the acetic acid \(=1.8 \times 10^{-5}\). (a) \(1200 \mathrm{ml}\) (b) \(300 \mathrm{ml}\) (c) \(600 \mathrm{ml}\) (d) \(900 \mathrm{ml}\)

The dissociation constant of formic acid is \(0.00024\). The hydrogen ion concentration in \(0.002 \mathrm{M}-\mathrm{HCOOH}\) solution is nearly (a) \(6.93 \times 10^{-4} \mathrm{M}\) (b) \(4.8 \times 10^{-7} \mathrm{M}\) (c) \(5.8 \times 10^{-4} \mathrm{M}\) (d) \(1.4 \times 10^{-4} \mathrm{M}\)

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