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\(100 \mathrm{ml}\) of \(0.015 \mathrm{M} \mathrm{HCl}\) solution is mixed with 100 \(\mathrm{ml}\) of \(0.005 \mathrm{M} \mathrm{HCl}\). What is the \(\mathrm{pH}\) of the resultant solution? (a) \(2.5\) (b) \(1.5\) (c) 2 (d) 1

Short Answer

Expert verified
The pH of the resultant solution is 2 (option c).

Step by step solution

01

Calculate Total Volume

When two solutions are mixed together, the total volume of the resultant solution is the sum of the individual volumes. Here, we combine 100 ml of 0.015 M HCl with 100 ml of 0.005 M HCl, thus the total volume becomes \(100 \text{ ml} + 100 \text{ ml} = 200 \text{ ml}\) or \(0.2\text{ L}\).
02

Determine Total Moles of HCl

Calculate the total moles of HCl in each solution. For the first solution: \( \text{moles} = 0.015 \text{ M} \times 0.1 \text{ L} = 0.0015 \text{ moles}\). For the second solution: \( \text{moles} = 0.005 \text{ M} \times 0.1 \text{ L} = 0.0005 \text{ moles}\). Sum these to find the total moles of HCl: \(0.0015 + 0.0005 = 0.002 \text{ moles}\).
03

Calculate Molarity of Resultant Solution

Use the formula \(M = \frac{\text{total moles}}{\text{total volume}}\) to calculate the molarity of the resultant solution. Here, \(M = \frac{0.002 \text{ moles}}{0.2 \text{ L}} = 0.01 \text{ M}\).
04

Calculate pH

The pH of a solution is calculated using the formula \(\text{pH} = -\log[\text{H}^+]\), where \([\text{H}^+]\) is the molarity of the HCl, which is 0.01 M as calculated. Therefore, \(\text{pH} = -\log(0.01) = 2\).
05

Conclusion: Match the Answer to Options

The calculated pH of the resultant solution is 2, which matches option (c).

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Key Concepts

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

HCl solution
Hydrochloric acid (HCl) is a strong acid commonly used in laboratories and industries. When we talk about an HCl solution, we are referring to hydrochloric acid dissolved in water. A characteristic feature of HCl and other strong acids is that they completely dissociate into ions when dissolved in water. This means every molecule of HCl breaks apart to form
  • Hydrogen ions (\(\text{H}^+\))
  • Chloride ions (\(\text{Cl}^-\))
This complete ionization makes HCl a strong acid, and it is classified accordingly because of its capacity to readily release hydrogen ions into the solution. The ionization in an aqueous solution is usually denoted as: \[\text{HCl (aq)} \rightarrow \text{H}^+ \text{(aq)} + \text{Cl}^- \text{(aq)}\] So, when we calculate the pH of an HCl solution, we are effectively determining the concentration of hydrogen ions in the solution.
molarity
Molarity is a way to express the concentration of a solution and is defined as the moles of solute per liter of solution. Knowing the molarity of a solution helps in understanding how "strong" or concentrated the solution is.
  • Molarity formula: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \)
  • A 1 M solution implies there is 1 mole of solute in 1 liter of solution.
In our exercise, we dealt with two solutions of HCl, each having different molarities:
  • 0.015 M and 0.005 M HCl.
By determining the moles of HCl in each solution, and then adding them together, we were able to calculate the molarity of the resultant solution after mixing. This calculation is crucial because it provides the foundation for further analysis, such as determining pH in acid-base reactions.
acid-base reactions
Acid-base reactions are fundamental chemical reactions where protons (\(\text{H}^+\) ions) are transferred between reactants. An acid, like HCl, donates protons, while a base typically accepts them. These reactions are essential not only in chemistry labs but also in everyday life, including digestion and buffering solutions that maintain pH levels in our body systems.In most cases, when you mix an acid with a base, they neutralize each other, producing water and a salt. However, in our specific exercise, we were dealing only with an acid solution. This means our primary focus was on calculating the free hydrogen ion concentration, which determines the solution's acidity, expressed by its pH.The pH scale is a numerical representation of how acidic or basic a solution is. It is logarithmic, meaning each whole number change on the scale represents a tenfold increase or decrease in hydrogen ion concentration. For our HCl solution, acquiring a pH involved calculating \(-\log[\text{H}^+]\)based on the total molarity of hydrogen ions in the mixed solution.

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Most popular questions from this chapter

For the reaction, \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) the equilibrium concentration of \(\mathrm{H}_{2}, \mathrm{I}_{2}\) and \(\mathrm{HI}\) are \(8.0,3.0\) and \(28.0\) mole/litre, respectively, the equilibrium constant is: (a) \(28.34\) (b) \(32.66\) (c) \(34.78\) (d) \(38.88\)

Le Chatelier's Principle is applicable to: (a) Heterogenous reaction (b) Homogenous reaction (c) Irreversible reaction (d) System in equilibrium

At \(700 \mathrm{~K}\), the equilibrium constant \(\mathrm{K}_{\mathrm{p}}\) for the reaction \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) is \(1.80 \times 10^{-3}\) What is the numerical value in mole per litre of equilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for this reaction at the same temperature: (a) \(8.1 \times 10^{-8}\) (b) \(9.1 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(3.1 \times 10^{-7}\) (d) \(6.1 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}+22.4 \mathrm{kcal}\) formation of \(\mathrm{NH}_{3}\) by above reaction shows: (a) Cyanamide process (b) Serpeck's process (c) Haber process (d) None of these

If equilibrium constants of reaction: \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) is \(\mathrm{K}_{1}\), and \(\frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) is \(\mathrm{K}_{2}\) then (a) \(\mathrm{K}_{1}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{1}=2 \mathrm{~K}_{2}\) (c) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}}_{1}\) (d) \(\mathrm{K}_{1}=\frac{1}{2} \mathrm{~K}_{2}\)

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