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How many liters of distilled water must be added to 1 liter of an aqueous solution of HCl with a pH of 1 to create a solution with a pH of 2? (A) 0.1 L (B) 0.9 L (C) 2 L (D) 9 L

Short Answer

Expert verified
The answer is (D) 9 L.

Step by step solution

01

Determine initial concentration of \(H_3O^+\) ions

From the definition of pH, we can determine the initial concentration, \(C_1\), of \(H_3O^+\) ions in the HCl solution by using the equation \(C_1=10^{-pH}\). So, when \(pH=1\), \(C_1=10^{-1}=0.1\) M (moles per liter).
02

Determine final concentration of \(H_3O^+\) ions

We will also aim for a final concentration, \(C_2\), of \(H_3O^+\) ions that would give a pH of 2. Similarly, we use the equation \(C_2=10^{-pH}\). So, when pH=2, \(C_2=10^{-2}=0.01\) M.
03

Calculate the volume of distilled water to be added

In order to reach the intended concentration, the same amount of substance (moles of \(H_3O^+\)) needs to be diluted with a total of 10 volumes (because \(C_1/C_2 = 10\)), which means adding 9 volumes of water to the initial one. Therefore, 9 liters of distilled water should be added to the solution.

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Key Concepts

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

Dilution
Dilution is a process where the concentration of a solute in a solution is reduced by adding more solvent. This allows us to create a solution of desired concentration by simply adding more of the solvent, often water, to the original solution. When we dilute a solution, the total number of solute particles remains the same, but they are spread out over a larger volume of solvent, leading to a decrease in concentration. In the context of the exercise, you start with a concentrated HCl solution and dilute it with distilled water to reduce its pH from 1 to 2. This involves making the hydronium ion concentration less, which corresponds to a more dilute acid solution.
HCl Solution
HCl, or hydrochloric acid, is a strong acid that dissociates completely in water. This means that when you have an HCl solution, it readily gives up its hydrogen ions (H+) to form hydronium ions ({\( H_3O^+ \) }).For instance, in the initial solution provided in the exercise, an HCl solution with a pH of 1 implies a very high hydronium ion concentration, specifically 0.1 M.Understanding how HCl behaves in solution helps in predicting how its acidity changes during dilution, as it directly relates to the change in concentration of {\( H_3O^+ \)} ions. With a reduction in pH, we observe a tenfold decrease in the hydronium ion concentration with each pH unit increase, indicative of the logarithmic nature of the pH scale.
Hydronium Ion Concentration
The hydronium ion concentration ({\( H_3O^+ \)}) essentially governs the acidity of a solution. It is closely linked to the pH of the solution, which is calculated using the formula:\[ pH = -\log[H_3O^+] \]From this equation, we can find out how acidic or basic a solution is based on the concentration of hydronium ions present. To convert a solution from one pH level to another, as in our exercise, you change this concentration. For example, going from pH 1 to pH 2 means reducing {\( H_3O^+ \)} by a factor of ten, lowering it from 0.1 M to 0.01 M.
Distilled Water
Distilled water is pure water that contains no dissolved salts, minerals, or impurities. It is often used as a solvent in chemical reactions because it does not introduce any additional ions into the solution. In dilution processes, such as those involving hydrochloric acid, distilled water is ideal because it does not alter the chemical composition of the solution other than decreasing the concentration of the solute. This property makes distilled water perfect for achieving precise dilutions, like those required to change the pH of the HCl solution from 1 to 2 by carefully adding 9 liters to dilute the initial concentration of hydronium ions.

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Most popular questions from this chapter

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. Identify the correct net ionic equation for the reaction that takes place. (A) \(\mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q) \mapsto \mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (B) \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \leftrightarrow \mathrm{NaCN}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (D) \(\mathrm{H}^{+}(a q)+\mathrm{CN}^{-}(a q)+\mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q)+\mathrm{Na}^{+}\) (aq)

$$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g)$$ When the reaction given above takes place in a sealed isothermal container, the rate law is $$\text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]$$ If a mole of \(\mathrm{H}_{2}\) gas is added to the reaction chamber and the temperature remains constant, which of the following will be true? (A) The rate of reaction and the rate constant will increase. (B) The rate of reaction and the rate constant will not change. (C) The rate of reaction will increase and the rate constant will decrease. (D) The rate of reaction will increase and the rate constant will not change.

\(2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.

Which of the substances would be soluble in water? (A) Ethylene glycol only, because it has the longest bond lengths (B) Acetone only, because it is the most symmetrical (C) Ethanol and ethylene glycol only, because of their hydroxyl (-OH) (D) All three substances would be soluble in water due to their permanent dipoles.

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ When the cell is connected, which of the following reactions takes place at the anode? (A) \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) (B) \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}+2 e^{-}\) (C) \(2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2}(g)\) (D) \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}+2 e^{-}\)

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