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At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\). Calculate \(\left[\mathrm{H}^{+}\right]\)and \(\left[\mathrm{OH}^{-}\right]\)for a neutral solution at this temperature.

Short Answer

Expert verified
In a neutral solution at 0掳C, the concentrations of H鈦 and OH鈦 ions are approximately \(1.09 \times 10^{-8}\) M.

Step by step solution

01

Write down the ion product constant of water equation for a neutral solution

The ion product constant of water equation is given by: \(K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\) Since we have a neutral solution, the concentrations of H鈦 and OH鈦 ions are equal: \(\left[\mathrm{H}^{+}\right] = \left[\mathrm{OH}^{-}\right] = x\) Now we can rewrite the equation as: \(K_w = x^2\)
02

Substitute the given value of \(K_w\) and solve for x

We are given the value of \(K_w\) at 0掳C: \(K_w = 1.2 \times 10^{-15}\) Now substitute this value into the equation and solve for x: \(1.2 \times 10^{-15} = x^2\) To solve for x, take the square root of both sides: \(x = \sqrt{1.2 \times 10^{-15}}\)
03

Calculate x's numeric value

Now let's calculate the numeric value of x: \(x \approx 1.09 \times 10^{-8}\)
04

Find the concentrations of H鈦 and OH鈦 ions

Since x is equal to the concentrations of H鈦 and OH鈦 ions in the neutral solution, we can write the final values as: \(\left[\mathrm{H}^{+}\right] = \left[\mathrm{OH}^{-}\right] \approx 1.09 \times 10^{-8} \, \mathrm{M}\) So, at 0掳C, in a neutral solution, the concentrations of H鈦 and OH鈦 ions are approximately \(1.09 \times 10^{-8}\) M.

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Key Concepts

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

Freezing Point
The freezing point is an essential concept when discussing the behavior of substances like water. At the freezing point, which for water is 0 degrees Celsius (32 degrees Fahrenheit), water begins to transition from a liquid to a solid state. This temperature is significant because the properties of water change and this also affects its chemical behavior.

For instance, the ion product constant of water, denoted as \( K_w \), depends on the temperature. At 0掳C, \( K_w \) is \( 1.2 \times 10^{-15} \), quite different from its value at 25掳C (which is \( 1.0 \times 10^{-14} \)). This implies that water behaves differently at its freezing point compared to higher temperatures.
  • Understanding the freezing point helps to predict how substances dissolve or ionize under different thermal conditions.
  • The freezing point is fundamental for processes such as crystallization and for understanding weather patterns.
Neutral Solution
A neutral solution is a special state where the concentration of hydrogen ions \( [\mathrm{H}^+] \) equals the concentration of hydroxide ions \( [\mathrm{OH}^-] \). This balance results in a solution that is neither acidic nor basic, typically understood as having a pH of 7 at standard conditions (25掳C).

However, this balance changes with temperature. At the freezing point (0掳C), the concentrations of ions are determined by the \( K_w \) value at that temperature. In a neutral solution at 0掳C, \( K_w = 1.2 \times 10^{-15} \), so both \( [\mathrm{H}^+] \) and \( [\mathrm{OH}^-] \) are approximately \( 1.09 \times 10^{-8} \, \text{M} \). This deviation from the classic \( 1.0 \times 10^{-7} \) M at room temperature is due to temperature's impact on ionization.
  • Neutral solutions help to understand varying pH conditions across different temperatures.
  • These principles are applicable in fields like biology and environmental science.
Hydrogen Ion Concentration
Hydrogen ion concentration, represented as \( [\mathrm{H}^+] \), is pivotal in determining the acidity of a solution. The concentration level signifies a solution's pH, which reflects its acidic or basic nature.

In the context of a neutral solution at the freezing point of water (0掳C), the calculation involves using the ion product constant of water \( K_w \). By solving the equation \( K_w = [\mathrm{H}^+][\mathrm{OH}^-] \), where both concentrations are equal, we find \( [\mathrm{H}^+] = 1.09 \times 10^{-8} \, \text{M} \).
  • This value affects how reactions proceed in biochemical and industrial processes under varying thermal conditions.
  • The concept is crucial in learning about buffers, neutralization, and water chemistry.

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Most popular questions from this chapter

Using data from Appendix \(\mathrm{D}\), calculate \(\left[\mathrm{OH}^{-}\right]\)and \(\mathrm{pH}\) for each of the following solutions: (a) \(0.105 \mathrm{M} \mathrm{NaF}\), (b) \(0.035 \mathrm{MNa}_{2} \mathrm{~S}\), (c) a mixture that is \(0.045 \mathrm{M}\) in \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(0.055 \mathrm{M}\) in \(\mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\).

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\)or \(\mathrm{NO}_{2}^{-}\), (b) \(\mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-}\), (c) \(\mathrm{HCO}_{3}^{-}\)or \(\mathrm{CO}_{3}^{2-}\).

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\)or \(\mathrm{ClO}^{-}\), (b) \(\mathrm{BrO}^{-}\)or \(\mathrm{BrO}_{2}^{-}\), (c) \(\mathrm{HPO}_{4}{ }^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\).

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 \mathrm{MHNO}_{3}\), (b) \(0.225 \mathrm{~g}\) of \(\mathrm{HClO}_{3}\) in \(2.00 \mathrm{~L}\) of solution, (c) \(15.00 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}\), (d) a mixture formed by adding \(50.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HI}\).

Pyridinium bromide \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)\) is a strong electrolyte that dissociates completely into $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\( and \)\mathrm{Br}^{-}$. An aqueous solution of pyridinium bromide has a pH of \(2.95 .\) (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the \(K_{a}\) for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation at equilibrium, in units of molarity?

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