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The \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) hydrochloric acid is not \(8.00\). The correct \(\mathrm{pH}\) can be calculated by considering the relationship between the molarities of the three principal ions in the solution \(\left(\mathrm{H}^{+}, \mathrm{Cl}^{-}\right.\), and \(\left.\mathrm{OH}^{-}\right) .\) These molarities can be calculated from algebraic equations that can be derived from the considerations given below. a. The solution is electrically neutral. b. The hydrochloric acid can be assumed to be \(100 \%\) ionized. c. The product of the molarities of the hydronium ions and the hydroxide ions must equal \(K_{\mathrm{w}}\). Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-8}-\mathrm{M} \mathrm{HCl}\) solution.

Short Answer

Expert verified
The pH of a \(1.0 \times 10^{-8}\) M HCl solution is approximately 6.0.

Step by step solution

01

Write the equation for the dissociation of HCl

The hydrochloric acid (HCl) dissociates completely as follows: \(HCl \rightarrow H^{+} + Cl^{-}\)
02

Write the statement for electrical neutrality

Since the solution is electrically neutral, the sum of the positive ions (H鈦) must equal the sum of negative ions (Cl鈦 and OH鈦). \[H^{+} = Cl^{-} + OH^{-}\]
03

Write the relation involving the auto-ionization constant of water (Kw)

The product of the concentrations of H鈦 and OH鈦 in water at 25鈩 equals the auto-ionization constant of water (Kw = 1.0 脳 10鈦宦光伌). \[H^{+} \times OH^{-} = 1.0 \times 10^{-14}\]
04

Substitute the given value of [HCl] in the electrical neutrality equation

From the complete ionization of HCl, [H鈦篯 becomes the same as [HCl], so: \[1.0 \times 10^{-8} = Cl^{-} + OH^{-}\]
05

Solve for [OH鈦籡 using the equations from Steps 3 and 4

Substitute [H鈦篯 from the electrical neutrality equation into the relation involving Kw: \[(1.0 \times 10^{-8}) \times OH^{-} = 1.0 \times 10^{-14}\] Solve for [OH鈦籡: \[OH^{-} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-8}}\] \[OH^{-} = 1.0 \times 10^{-6}\]
06

Calculate the concentration of H鈦 using the relation obtained in Step 4

Substitute the concentration of [OH鈦籡 into the electrical neutrality equation: \[1.0 \times 10^{-8} = Cl^{-} + (1.0 \times 10^{-6})\] Solve for [Cl鈦籡, which equals [H鈦篯 due to complete ionization of HCl: \[H^{+} = 1.0 \times 10^{-8} - 1.0 \times 10^{-6}\] \[H^{+} = 9.9 \times 10^{-7}\]
07

Calculate the pH of the solution

Finally, use the formula pH = -log[H鈦篯 to calculate the pH of the solution: \[pH = -\log (9.9 \times 10^{-7})\] \[pH \approx 6.0\] Hence, the pH of a \(1.0 \times 10^{-8}\) M HCl solution is approximately 6.0.

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Key Concepts

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

Hydrochloric Acid Dissociation
Hydrochloric acid (HCl) is a strong acid, which means it dissociates completely in water. This complete dissociation allows it to break down into hydrogen ions (H鈦) and chloride ions (Cl鈦) without leaving any undissociated HCl molecules in the solution. The dissociation reaction is represented as:
  • \[ HCl \rightarrow H^{+} + Cl^{-} \]
This characteristic of HCl is important when calculating the pH, as the concentration of H鈦 ions directly equals the initial concentration of HCl. Thus, it becomes crucial to consider both the H鈦 ions from the dissociation of HCl and the auto-ionization of water. By understanding this fundamental property of hydrochloric acid, we can accurately predict and calculate the pH of its solution, even at such low concentrations.
Auto-Ionization Constant of Water (Kw)
Water slightly ionizes into hydrogen ions (H鈦) and hydroxide ions (OH鈦), a process known as auto-ionization. The product of the concentrations of these ions is constant at a particular temperature (especially at 25鈩) and is given by the expression:
  • \[ H^{+} \times OH^{-} = K_{w} = 1.0 \times 10^{-14} \]
This constant, known as the auto-ionization constant of water, is crucial when dealing with very dilute acid or base solutions, like in the case of 1.0 x 10鈦烩伕 M HCl. Since the concentration of H鈦 from HCl is comparable to the concentration from water, you must account for both sources when calculating the final pH. Therefore, understanding that Kw provides a relationship between H鈦 and OH鈦 ion concentrations helps manage these calculations in acid-base equilibrium problems.
Electrical Neutrality of Solution
In any solution, the total charge should balance to maintain electrical neutrality. For acidic solutions like hydrochloric acid, this involves balancing the concentrations of positive ions, primarily H鈦, and negative ions such as Cl鈦 and OH鈦.
  • The principle can be expressed as: \[ H^{+} = Cl^{-} + OH^{-} \]
This equation ensures that the sum of positive charges equals the sum of negative charges. In the context of the HCl solution, the H鈦 ions come from both the dissociation of HCl and the auto-ionization of water. Negative ions come from Cl鈦 produced from HCl dissociation and OH鈦 from water. Achieving electrical neutrality is crucial for accurately calculating the \mathrm{pH} in dilute solutions, motivating consideration of all ion species in these calculations.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a \(0.050-\mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.4 \times 10^{-5}\).

The \(\mathrm{pH}\) of a \(0.063-M\) solution of hypobromous acid \((\mathrm{HOBr}\) but usually written \(\mathrm{HBrO}\) ) is 4.95. Calculate \(K_{\mathrm{a}}\).

Place the species in each of the following groups in order of increasing base strength. Give your reasoning in each case. a. \(\mathrm{IO}_{3}^{-}, \mathrm{BrO}_{3}^{-}\) b. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\) c. \(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\)

What are the major species present in \(0.015 \mathrm{M}\) solutions of each of the following bases? a. \(\mathrm{KOH}\) b. \(\mathrm{Ba}(\mathrm{OH})_{2}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

A sample containing \(0.0500\) mole of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is dissolved in enough water to make \(1.00 \mathrm{~L}\) of solution. This solution contains hydrated \(\mathrm{SO}_{4}{ }^{2-}\) and \(\mathrm{Fe}^{3+}\) ions. The latter behaves as an acid: $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)$$ a. Calculate the expected osmotic pressure of this solution at \(25^{\circ} \mathrm{C}\) if the above dissociation is negligible. b. The actual osmotic pressure of the solution is \(6.73\) atm at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\). (To do this calculation, you must assume that none of the ions go through the semipermeable membrane. Actually, this is not a great assumption for the tiny \(\mathrm{H}^{+}\) ion.)

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