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Chapter 5: Problem 69

\(100 \mathrm{cc}\) of \(\mathrm{HCl}\) of \(\mathrm{pH}\) value 1 is mixed with \(100 \mathrm{cc}\) of distilled water. The \(\mathrm{pH}\) of the resultant solution is a. \(1.3\) b. \(1.7\) c. \(2.3\) d. \(1.9\)

Short Answer

Expert verified
The pH of the resultant solution is 1.3.

Step by step solution

01

Determine Initial Molarity of HCl

We are given that the pH of HCl solution is 1. By definition, \( \text{pH} = -\log[\text{H}^+] \), which means that the concentration of hydrogen ions, \( [\text{H}^+] \), is \( 10^{-1} = 0.1 \) M (molarity).
02

Calculate the Moles of HCl

The volume of the HCl solution is 100 cc (or 0.1 L). Using the molarity, the moles of HCl is found using \( \text{moles} = \text{molarity} \times \text{volume} \). Hence, \( 0.1 \text{ M} \times 0.1 \text{ L} = 0.01 \text{ moles of HCl} \).
03

Calculate Total Volume after Mixing

After mixing 100 cc of HCl solution with 100 cc of distilled water, the total volume becomes \( 100 \text{ cc} + 100 \text{ cc} = 200 \text{ cc} = 0.2 \text{ L} \).
04

Calculate New Molarity of HCl

The moles of HCl remain constant; only the volume changes, impacting the molarity. The new molarity is \( \text{moles of HCl} / \text{total volume} = 0.01 / 0.2 = 0.05 \text{ M} \).
05

Calculate New pH of the Solution

Using the relation \( \text{pH} = -\log[\text{H}^+] \) once again, substitute \([\text{H}^+] = 0.05 \text{ M}\). Therefore, \( \text{pH} = -\log(0.05) \approx 1.3 \).

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Key Concepts

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

Molarity
Molarity is a fundamental concept in chemistry. It defines the concentration of a solute in a solution, measured in moles per liter (M). When we talk about calculating the molarity of a solution, we're essentially determining how many moles of a substance are present in one liter of the solution. This concept is crucial in various calculations such as dilution and reactions.
To find molarity, you use the formula:
  • Molarity (\( M \)) = Moles of solute / Volume of solution (L)
In the exercise provided, the initial pH of the Hydrochloric Acid (HCl) solution was 1. This information directly gives us the hydrogen ion concentration, which is equal to 0.1 M, or \( 10^{-1} \) M. With the volume given as 100 cubic centimeters (cc), we converted the units to liters (0.1 L) to align them with the molarity formula. Hence, the molarity of the HCl solution before dilution was calculated to be 0.1 M.
Hydrogen Ion Concentration
The hydrogen ion concentration \([ ext{H}^+]\) in a solution is pivotal in determining its acidity or basicity, represented by its pH. The pH is calculated using the formula:
  • \( ext{pH} = - ext{log}[ ext{H}^+]\)
A lower pH indicates a higher concentration of hydrogen ions, meaning stronger acidity. For instance, in our problem, the initial concentration was found to be 0.1 M, correlating with a pH of 1. This indicates a highly acidic solution.
During dilution or mixing, even though the total volume changes, the moles of hydrogen ions stay the same. The hydrogen ion concentration changes because it is now spread across a larger volume, impacting the resultant solution's pH. After dilution, the concentration decreased, leading to a new pH value. Understanding these changes and how they affect pH is key to mastering acid-base chemistry.
Solution Dilution
Dilution is the process of reducing the concentration of a solute in a solution. This is done by adding more solvent to it. When mixing solutions, the number of solute particles remains constant while the volume increases. This causes the concentration to decrease.
In our example problem, we diluted the HCl solution by doubling its volume, mixing 100 cc of HCl with 100 cc of distilled water. The new volume was 200 cc or 0.2 L, while the moles of HCl stayed fixed at 0.01 moles. Calculating the new molarity, we found it to be 0.05 M, showing how dilution affects the concentration: M = Moles of solute / New total volume.
Since pH is a direct function of hydrogen ion concentration, and dilution decreases concentration, the pH rose from 1 to approximately 1.3. As seen here, pH calculations illustrate how crucial it is to consider both initial conditions and dilution factors.

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Most popular questions from this chapter

\(\mathrm{Ag}^{+}+\mathrm{NH}_{3} \leftrightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)^{+}\right]\) \(\mathrm{K}_{1}=3.5 \times 10^{-3}\) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)\right]^{+}+\mathrm{NH}_{3} \leftrightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} ;\) \(\mathrm{K}_{2}=1.7 \times 10^{-3}\) Then the formation constant of \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\)is a. \(6.08 \times 10^{-6}\) b. \(6.08 \times 10^{6}\) c. \(6.08 \times 10^{-9}\) d. None

When \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (ionization constant \(\left.\mathrm{K}_{\mathrm{b}}=5 \times 10^{-4}\right)\) is mixed with \(0.08 \mathrm{~mol} \mathrm{HCl}\) and the volume is made up of 1 litre. Find the \(\left[\mathrm{H}^{+}\right]\)of resulting solution. a. \(8 \times 10^{-2}\) b. \(2 \times 10^{-11}\) c. \(1.23 \times 10^{-4}\) d. \(8 \times 10^{-11}\)

Which of the following are homogeneous equilibria? a. \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) b. \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{X}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HX}(\mathrm{g})\) c. \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) d. \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

The hexaammine cobalt (III) ion is very unstable in acidic aqueous solution: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}(\mathrm{aq})+6 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow\) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{4+}(\mathrm{aq})+6 \mathrm{NH}_{4}^{+}(\mathrm{aq})\) However, solutions of hexaammine cobalt(III) can be stored in acidic solution for months without noticeable decomposition. Which statement below about the equilibrium constant and the activation energy for the reaction is true? a. \(\mathrm{Keq}>10^{3}\) and \(\mathrm{Ea}\) is very large b. Keq \(<10^{3}\) and Ea is very small c. Keq \(<10^{3}\) and Ea is very large d. \(\mathrm{Keq}>10^{3}\) and Ea is very small

Which of the following statement is correct? 1\. the \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) solution of \(\mathrm{HCl}\) is 8 2\. the conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{HPO}_{4}^{2-}\) 3\. autoprotolysis constant of water increases with temperature 4\. when a solution of a weak monoprotic acid is titrated against a strong base at half neutralization point, \(\mathrm{pH}=(1 / 2) \mathrm{pKa}\). a. 2,3 b. \(1,2,3\) c. 3,4 d. \(2,3,4\)
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