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Chapter 7: Problem 12

\(V_{a} \mathrm{~mL}\) of a strong acid solution of \(\mathrm{pH} 2.00\) is mixed with \(V_{b} \mathrm{~mL}\) of a strong base solution of \(\mathrm{pH}\) 11.00. Express \(V_{a}\) in terms of \(V_{b}\) if the mixture is neutral. The solution temperature is \(24^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Va = Vb

Step by step solution

01

Calculate the Hydrogen Ion Concentration for the Acid

The \(\text{pH}\) of a solution is given by the expression \[\text{pH} = -\log_{10}[H^+]\], where \[H^+]\] represents the concentration of hydrogen ions in mol/L. Given that the \(\text{pH}\) of the strong acid solution is 2.00, we can find \[H^+]\] as follows:\[H^+] = 10^{-\text{pH}} = 10^{-2} = 0.01 \text{mol/L}\]

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Key Concepts

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

pH calculation
Understanding how to calculate the pH of a solution is essential in acid-base chemistry. The term "pH" refers to the potential of hydrogen in the solution and is a measure used to determine how acidic or basic a solution is. The pH scale ranges from 0 to 14, where a pH of 7 is neutral, below 7 is acidic, and above 7 is basic.
To calculate the pH of a solution, the formula \[ \text{pH} = -\log_{10}[H^+] \]is used. Here, \([H^+]\) denotes the concentration of hydrogen ions in the solution in mol/L.
A strong acid or base will completely dissociate in water, drastically affecting the pH. Remember, small changes in pH can represent large changes in \([H^+]\), due to the logarithmic scale used for pH calculation.
hydrogen ion concentration
Hydrogen ion concentration is a crucial concept when discussing the acidity or basicity of a solution. It is typically expressed in terms of molarity, (mol/L), and determines how acidic or alkaline a solution is.
The concentration of hydrogen ions, \([H^+]\), is inversely correlated with the pH of a solution, due to the formula\[ \text{pH} = -\log_{10}[H^+] \]. As the concentration of hydrogen ions increases, the pH value decreases, indicating a more acidic solution. Conversely, as \([H^+]\) decreases, the pH increases, leading to a more basic solution.
In our given example, the strong acid's pH is 2.00, which implies a hydrogen ion concentration of 0.01 mol/L, or \(10^{-2}\). Understanding these measurements is key to grasping how acids and bases neutralize each other.
strong acid and strong base
Strong acids and bases are substances that completely dissociate into ions in water. This characteristic makes them significantly alter the pH of a solution.
A strong acid releases a large number of \(H^+\) ions when dissolved in water, while a strong base releases hydroxide ions \((OH^-)\). Because these compounds dissociate completely, they are an important part of chemical reactions, particularly neutralization.
Neutralization occurs when an acid and base react to form water and a salt, effectively cancelling out each other's characteristics. In the exercise, when a strong acid solution of pH 2.00 is mixed with a strong base solution of pH 11.00, the goal is to achieve a neutral mixture with a pH of 7.00. Understanding how strong acids and bases interact helps explain how different volumes of each are required to reach neutrality.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of the solution prepared by adding \(0.10 \mathrm{~mol}\) each of hydroxylamine and hydrochloric acid to \(500 \mathrm{~mL}\) water.

\(K_{a}\) for acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is \(1.75 \times 10^{-5} . K_{w}\) is \(1.00 \times 10^{-14}\) (a) Find \(K_{b}\) for acetate ion\(\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)\) (b) When \(0.1 M\) of sodium acetate \(\left(\mathrm{CH}_{3}\right.\) COONa \()\) dissolves in water at \(24^{\circ} \mathrm{C}\), what is the \(\mathrm{pH}\) of the solution? Assume the ions behave ideally.

The \(\mathrm{pH}\) of a \(0.20 \mathrm{M}\) solution of a primary amine, \(\mathrm{RNH}_{2},\) is \(8.42 .\) What is the \(\mathrm{p} K_{b}\) of the amine?

An acetic acid-sodium acetate buffer of \(\mathrm{pH} 5.00\) is \(0.100 \mathrm{M}\) in \(\mathrm{NaOAc}\). Calculate the \(\mathrm{pH}\) after the addition of \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) to \(100 \mathrm{~mL}\) of the buffer.

Many geochemical processes are governed by simple chemical equilibria. One example is the formation of stalactites and stalagmites in a limestone cave, and is a good illustration of Henry's law. This is illustrated in the diagram below: Rainwater percolates through soil. Due to microbial activity in the soil, the gaseous \(\mathrm{CO}_{2}\) concentration in the soil interstitial space (expressed as the partial pressure of \(\mathrm{CO}_{2}, \mathrm{pCO}_{2}\). in atmospheres), is \(3.2 \times 10^{-2}\) atm, significantly higher than that in the ambient atmosphere \(\left(3.9 \times 10^{-4} \mathrm{~atm} ; \mathrm{CO}_{2}\right.\) concentration in the ambient atmosphere is presently increasing by \(2 \times 10^{-6}\) atm each year, see http://CO2now.org for the current atmospheric \(\mathrm{CO}_{2}\) concentration). Water percolating through the soil reaches an equilibrium (called Henry's law equilibrium) with the soil interstitial \(\mathrm{pCO}_{2}\) as given by Henry's law: $$ \left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]=K_{\mathrm{H}} \mathrm{pCO}_{2}$$ where \(\mathrm{H}_{2} \mathrm{CO}_{3}\) is the aqueous carbonic acid concentration and \(K_{\mathrm{H}}\) is the Henry's law constant for \(\mathrm{CO}_{2}, 4.6 \times 10^{-2} \mathrm{M}\) /atm at the soil temperature of \(15^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) -saturated water effluent from the soil layer then percolates through fractures and cracks in a limestone layer, whereupon it is saturated with \(\mathrm{CaCO}_{3} .\) This \(\mathrm{CaCO}_{3}\) saturated water drips from the ceiling of the cave. Because of the diurnal temperature variation outside the cave, the cave "breathes": the \(\mathrm{CO}_{2}\) concentration in the cave atmosphere is essentially the same as in ambient air \(\left(3.9 \times 10^{-4} \mathrm{~atm}\right)\) Show that when the water dripping from the ceiling re-equilibrates with the \(\mathrm{pCO}_{2}\) concentration in the cave atmosphere, some of the calcium in the drip water will re-precipitate as \(\mathrm{CaCO}_{3}\), thus forming stalactites and stalagmites. Assume cave temperature to be \(15^{\circ} \mathrm{C}\) as well. At this temperature the successive dissociation constants of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) are: \(K_{a 1}=3.8 \times 10^{-7}\) and \(K_{a 2}=3.7 \times 10^{-11}, K_{w}\) is \(4.6 \times 10^{-15}\) and \(K_{\mathrm{sp}}\) of \(\mathrm{CaCO}_{3}\) is \(4.7 \times 10^{-9}\) See the text website (as well as the Solutions Manual) for a detailed solution of this complex problem. Corresponding Goal Seek calculations are also given on the website.
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