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Chapter 2: Problem 8

Calculation of \(\mathrm{pH}\) from Concentration of Strong Acid Calculate the \(\mathrm{pH}\) of a solution prepared by diluting \(3.0 \mathrm{~mL}\) of \(2.5 \mathrm{M} \mathrm{HCl}\) to a final volume of \(100 \mathrm{~mL}\) with \(\mathrm{H}_{2} \mathrm{O}\).

Short Answer

Expert verified
The pH of the diluted solution is approximately 1.12.

Step by step solution

01

Calculate the Moles of HCl

First, we need to find the moles of HCl in the initial solution. We have a concentration of \(2.5 \, \text{M}\) and a volume of \(3.0 \, \text{mL}\). To find moles, use \(\text{moles} = \text{concentration} \times \text{volume in liters}\).\[\text{Moles of HCl} = 2.5 \, \text{M} \times 0.003 \, \text{L} = 0.0075 \, \text{mol}\]
02

Dilution Calculation

Next, we find the concentration of HCl after dilution to \(100 \, \text{mL}\). The number of moles remains constant during dilution, so:\[\text{New concentration} = \frac{0.0075 \, \text{mol}}{0.100 \, \text{L}} = 0.075 \, \text{M}\]
03

Calculate pH of Diluted Solution

Finally, we find the \(\text{pH}\) using the concentration of \(\text{HCl}\), which is a strong acid that dissociates completely in water.\(\text{pH} = -\log_{10}[\text{H}^+ ] = -\log_{10}([\text{HCl}] )\)\[\text{pH} = -\log_{10}(0.075) \approx 1.12\]

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Key Concepts

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

Strong Acid
When we talk about strong acids, we're referring to acids that completely dissociate in water. Hydrochloric acid (\(\text{HCl}\)) is one such strong acid. This means that every molecule of \(\text{HCl}\) added to water breaks apart into \(\text{H}^+\) and \(\text{Cl}^-\) ions.

Because strong acids dissociate entirely, the concentration of \(\text{H}^+\) ions is equal to the concentration of the acid. This characteristic makes \(\text{HCl}\) very predictable when calculating pH since the pH is determined solely by the initial molar concentration of the acid.

For weak acids, you would have to consider the equilibrium state where not all molecules dissociate. This often involves using an equilibrium constant (K) and can get quite complex. But with strong acids, like \(\text{HCl}\), things are a bit more straightforward; once you know the concentration, you can swiftly move to calculate the pH through the negative logarithm function.
Dilution
When you dilute a solution, you're adding more solvent (usually water) to it, which decreases the concentration of the solute. In our exercise, we diluted \(3.0 \, \text{mL}\) of a \(2.5 \, \text{M}\) \(\text{HCl}\) solution to a final volume of \(100 \, \text{mL}\).

The key aspect of dilution is that the number of moles of the solute (in this case, \(\text{HCl}\)) remains constant before and after the dilution.

  • Before dilation: 0.0075 moles in 3.0 mL
  • After dilution: same 0.0075 moles in 100 mL

This constancy allows you to calculate the new, lower concentration using the formula: \(\text{Concentration after dilution} = \frac{\text{moles}}{\text{new volume in liters}}\). This process is especially useful when you need less concentrated solutions for experimentation or application.
Molar Concentration
Molar concentration, often referred to as molarity, is a way of expressing the concentration of a solution. It tells us how many moles of a substance are present in one liter of solution.

For our \(\text{HCl}\) example, the initial molarity was given as \(2.5 \, \text{M}\). Here's how it works: \(\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\). So, if we know our solution's molarity and the volume, we can easily find out the number of moles of solute.

During dilution, the molarity went from \(2.5 \, \text{M}\) down to \(0.075 \, \text{M}\) after we added more water to extend the volume from \(3\, \text{mL}\) to \(100\, \text{mL}\).
  • This drop is due to the increased volume which spreads the same amount of solute over a larger amount of solvent.
  • After dilution, we used the new molarity to calculate the pH because strong acids like \(\text{HCl}\) dissociate entirely in water.
Ultimately, understanding molarity helps in accurately predicting pH levels, crucial in many scientific fields like chemistry and biology.

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Most popular questions from this chapter

pH and Drug Absorption Asp?rin is a weak acid with a \(p K_{n}\) of \(3.5\) (the ionizable \(H\) is shown in red): Aspirin is absorbed into the blood through the cells lining the stomach and the small intestine. Absorption requires passage through the plasma membrane. The polarity of the molecule determines the absorption rate: charged and highly polar molecules pass slowly, whereas neutral hydrophobic molecules pass rapidly. The \(\mathrm{pH}\) of the stomach contents is about \(1.5\), and the \(\mathrm{pH}\) of the contents of the small intestine is about 6. Rased on this information, is more aspirin absorbed into the bloodstream from the stomach or from the small intestine? Clearly justify your choice

Preparation of an Acetate Buffer Calculate the concentrations of acetic acid \(\left(\mathrm{p} K_{\mathrm{n}}-4.76\right)\) and sodium acetate necessary to prepare a \(0.2\) m buffer solution at pH \(5.0\).

Calculation of the \(\mathrm{pH}\) of a Strong Acid or Base a. Write out the acid dissociation reaction for hydrochloric acid. b. Calculate the \(\mathrm{pH}\) of a solution of \(5 \times 10^{-4} \mathrm{M}\) hydrochloric acid at \(25^{\circ} \mathrm{C}\). c. Write out the acid dissociation reaction for sodium hydroxide. d. Calculate the \(\mathrm{pH}\) of a solution of \(7 \times 10^{-5} \mathrm{M}\) sodium hydroxide at \(25^{\circ} \mathrm{C}\).

Choice of Weak Acid for a Buffer Determine whether each weak acid would best buffer at \(\mathrm{pH} 3.0\), at \(\mathrm{pH} 5.0\), or at \(\mathrm{pH} 9.0\) : a. formic acid \(\left(p K_{\mathrm{x}}=3.8\right)\); b. acetic acid \(\left(p K_{a}=4.76\right)\); c. ammonium \(\left(\mathrm{p} K_{\mathrm{n}}-9.25\right) ;\) d. boric acid \(\left(\mathrm{p} K_{\mathrm{a}}=9.24\right)\); e. chloroscetic acid \(\left(\mathrm{p} K_{\mathrm{z}}=2.87\right)\); f. hycdrazoic acid \(\left(p K_{a}=4.6\right)\). Briefly justify your answer.

Solubility of Ethanol in Water Ethane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{3}\right)\) and ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) differ in their molecular makeup by only one atom, yet ethanol is much more soluble in water than ethane. Describe the features of ethanol that make it more water soluble than ethane.
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