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What is the sum of the \(\mathrm{pH}\) and the \(\mathrm{pOH}\) for a solution at \(37^{\circ} \mathrm{C}\), where \(K_{w}\) equals \(2.5 \times 10^{-14} ?\)

Short Answer

Expert verified
The sum of pH and pOH at \(37^{\circ}C\) is 13.602.

Step by step solution

01

Understand the Relationship

The sum of the pH and pOH of a solution is determined by the ion product of water \(K_w\). At any temperature, \(K_w = [H^+][OH^-]\). The relationship between pH, pOH, and \(K_w\) is defined by the equation: \(\text{pH} + \text{pOH} = -\log(K_w)\).
02

Calculate the Logarithm

Given that \(K_w = 2.5 \times 10^{-14}\) at \(37^{\circ}C\), we need to compute \(-\log(2.5 \times 10^{-14})\) to find the sum of pH and pOH.
03

Apply Logarithmic Properties

Use the property of logarithms \(-\log(a \times 10^b) = -\log(a) - b\log(10)\). Here \(a = 2.5\) and \(b = -14\). Compute \(-\log(2.5)\) and add \(14\) to find the result.
04

Calculate -Log(2.5)

\(-\log(2.5)\approx -0.398\). Adding this to 14 (the negative of the exponent in the base 10) results in \(14 - 0.398 = 13.602\). This value is \(-\log(2.5 \times 10^{-14})\).
05

Present Final Sum

Thus, the sum of the pH and the pOH of the solution at \(37^{\circ}C\) is \(13.602\).

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Key Concepts

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

Ion Product of Water
The concept of the ion product of water, represented as \(K_w\), is fundamental in understanding the behavior of acids and bases in aqueous solutions. Water has a very special ability to undergo a process called autoionization. This means it can dissociate into hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\). This dissociation can be represented by the equilibrium equation:
  • \(H_2O \rightleftharpoons H^+ + OH^-\)
At a given temperature, the product of the concentrations of these ions is a constant, \(K_w\). For pure water at 25°C, \(K_w = 1.0 \times 10^{-14}\). This value represents the limit of the product of \([H^+]\) and \([OH^-]\) even in the presence of other solutes.
Understanding \(K_w\) is crucial when calculating pH and pOH in solutions since it provides a relationship between these values. The sum of pH and pOH is equal to the negative logarithm of \(K_w\). This relationship is essential in determining the acidity or basicity of solutions.
Kw at Different Temperatures
The ion product of water, \(K_w\), is temperature-dependent, meaning its value changes with temperature. As temperature increases, the tendency of water molecules to dissociate into \(H^+\) and \(OH^-\) generally increases. This change in \(K_w\) is due to the shifting equilibrium according to Le Chatelier's Principle.
  • At 25°C, \(K_w = 1.0 \times 10^{-14}\)
  • At 37°C, as given in the problem, \(K_w = 2.5 \times 10^{-14}\)
When analyzing solutions at temperatures different from 25°C, it is important to use the corresponding \(K_w\) value. This influences both mathematical evaluations and conceptual understanding. Since the sum of pH and pOH relies on \(K_w\), deviations from standard conditions (25°C) necessitate recalculations.
This adjustment demonstrates how environmental conditions can affect chemical reactions, and particularly how temperature plays a critical role in aqueous equilibria.
Logarithmic Properties in Chemistry
Logarithms are a powerful tool in chemistry, particularly when dealing with phenomena such as acidity and basicity. The pH scale, which measures acidity, is itself a logarithmic scale. This means a change in one pH unit corresponds to a tenfold change in hydrogen ion concentration.
In calculations involving \(K_w\), understanding logarithmic properties is essential. For example, when determining the sum of pH and pOH, we utilize the property:
  • \(-\log(a \times 10^b) = -\log(a) - b\log(10)\)
This allows for precise calculations, as seen when \(K_w\) deviates from its standard value. Using \(-\log(2.5 \times 10^{-14})\), one can simplify calculations by considering:
  • \(-\log(2.5)\approx -0.398\)
  • The negative of the exponent, \(-(-14)\), adds 14 to this calculation
The application of these logarithmic properties allows chemists to navigate complex concepts with ease, simplifying the comparison and analysis of chemical states and reactions.

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Most popular questions from this chapter

Coal and other fossil fuels usually contain sulfur compounds that produce sulfur dioxide, \(\mathrm{SO}_{2}\), when burned. One possible way to remove the sulfur dioxide is to pass the combustion gases into a tower packed with calcium oxide, \(\mathrm{CaO}\). Write the equation for the reaction. Identify each reactant as either a Lewis acid or a Lewis base. Explain how you arrived at your answer.

You make a solution by dissolving \(0.0010\) mol of \(\mathrm{HCl}\) in enough water to make \(1.0 \mathrm{~L}\) of solution. a. Write the chemical equation for the reaction of \(\mathrm{HCl}(a q)\) and water. b. Without performing calculations, give a rough estimate of the \(\mathrm{pH}\) of the \(\mathrm{HCl}\) solution. Justify your answer. c. Calculate the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration and the \(\mathrm{pH}\) of the solution. d. Is there any concentration of the base \(\mathrm{OH}^{-}\) present in this solution of \(\mathrm{HCl}(a q) ?\) If so, where did it come from? e. If you increase the OH \(^{-}\) concentration of the solution by adding \(\mathrm{NaOH}\), does the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration change? If you think it does, explain why this change occurs and whether the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration increases or decreases. f. If you were to measure the \(\mathrm{pH}\) of 10 drops of the original HCl solution, would you expect it to be different from the pH of the entire sample? Explain. g. Explain how two different volumes of your original \(\mathrm{HCl}\) solution can have the same \(\mathrm{pH}\) yet contain different moles of \(\mathrm{H}_{3} \mathrm{O}^{+}\) h. If \(1.0 \mathrm{~L}\) of pure water were added to the \(\mathrm{HCl}\) solution, would this have any impact on the \(\mathrm{pH}\) ? Explain.

For each of the following, state whether the solution at \(25^{\circ} \mathrm{C}\) is acidic, neutral, or basic: (a) A \(0.1 M\) solution of trisodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4}\), has a pH of \(12.0 .\) (b) \(\mathrm{A} 0.1 \mathrm{M}\) solution of calcium chloride, \(\mathrm{CaCl}_{2}\), has a pH of \(7.0\). (c) A \(0.2 M\) solution of copper(II) sulfate, \(\mathrm{CuSO}_{4}\), has a pH of 4.0. (d) A sample of rainwater has a pH of \(5.7\).

The \(\mathrm{pH}\) of a solution is \(9.55\) at \(25^{\circ} \mathrm{C}\). What is the hydroxide-ion concentration in the solution? a. \(2.8 \times 10^{-10} M\) b. \(3.5 \times 10^{-5} M\) c. \(3.1 \times 10^{-10} M\) d. \(3.2 \times 10^{-5} M\) e. \(3.8 \times 10^{-5} M\)

Calculate the hydronium-ion concentration at \(25^{\circ} \mathrm{C}\) in a \(1.3 \times 10^{-2} M \mathrm{Ba}(\mathrm{OH})_{2}\) solution. a. \(1.3 \times 10^{-2} M\) b. \(7.7 \times 10^{-13} M\) c. \(2.6 \times 10^{-2} M\) d. \(3.8 \times 10^{-13} M\) e. \(1.5 \times 10^{2} M\)

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