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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°C with \(K_w = 1.2 \times 10^{-15}\), both the hydrogen ion concentration and the hydroxide ion concentration are approximately \(3.5 \times 10^{-8} M\).

Step by step solution

01

Write the Ion Product of Water Expression

For a neutral solution at 0°C, the ion product of water, \(K_w\), is given as \(1.2 \times 10^{-15}\). We can write the expression for the ion product of water as follows: \[K_w = \left[\mathrm{H}^{+}\right]^2\] Here, \(K_w\) is the ion product constant for water. The concentrations of hydrogen ions and hydroxide ions are equal in a neutral solution, since the number of hydrogen ions and the number of hydroxide ions are equal.
02

Solve for the Hydrogen Ion Concentration

Plug in the given value for \(K_w\) and solve for \(\left[\mathrm{H}^{+}\right]\): \(1.2\times 10^{-15} = \left[\mathrm{H}^{+}\right]^2\) To find the hydrogen ion concentration, take the square root of both sides: \(\left[\mathrm{H}^{+}\right] = \sqrt{1.2\times 10^{-15}}\)
03

Calculate the Hydrogen Ion Concentration

Evaluate the square root: \(\left[\mathrm{H}^{+}\right] \approx 3.5 \times 10^{-8} M\)
04

Find the Hydroxide Ion Concentration

Since the solution is neutral, the hydroxide ion concentration equals the hydrogen ion concentration: \(\left[\mathrm{OH}^{-}\right] = \left[\mathrm{H}^{+}\right]\) Thus, \(\left[\mathrm{OH}^{-}\right] \approx 3.5 \times 10^{-8} M\). In conclusion, for a neutral solution at 0°C, both the hydrogen ion concentration and the hydroxide ion concentration are approximately \(3.5 \times 10^{-8} M\).

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Key Concepts

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

Neutral Solution
In chemistry, a neutral solution is one where the concentrations of hydrogen ions (\(\left[\mathrm{H}^{+}\right]\)) and hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\)) are equal.This balance of ions results in a solution that is neither acidic nor basic.At different temperatures, the ion product of water, \(K_w\), varies, which affects the exact concentration of these ions in neutral solutions.
For example, at 0°C, which is the freezing point of water, the value of \(K_w\) is \(1.2 \times 10^{-15}\).This is different from the value at 25°C (\(1.0 \times 10^{-14}\)), which is more commonly used in room temperature conditions. Understanding how \(K_w\) changes with temperature helps predict the behavior of ions in a solution as conditions vary.
In practice, when dealing with neutral solutions at different temperatures, always remember:
  • \([\mathrm{H}^{+}] = [\mathrm{OH}^{-}]\)
  • The product \([\mathrm{H}^{+}][\mathrm{OH}^{-}] = K_w\)
This knowledge helps in calculating ion concentrations within neutral solutions.
Hydrogen Ion Concentration
The hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) is a measure of the acidity of a solution.In a neutral solution at any given temperature, \(\left[\mathrm{H}^{+}\right]\) is equal to the hydroxide ion concentration.To determine this concentration at a specific temperature, such as 0°C, we use the ion product of water, \(K_w\).
Given the equation \(K_w = \left[\mathrm{H}^{+}\right]^2\) for a neutral solution, solving for \(\left[\mathrm{H}^{+}\right]\) involves taking the square root of \(K_w\).For instance, at 0°C, \(K_w = 1.2\times 10^{-15}\), so we find:\[\left[\mathrm{H}^{+}\right] = \sqrt{1.2\times 10^{-15}}\]Evaluating this expression gives \(\left[\mathrm{H}^{+}\right] \approx 3.5 \times 10^{-8} \) M.This calculation reveals that the concentration of hydrogen ions in a neutral solution varies with temperature,underlining the importance of knowing \(K_w\) for precise acidity measurement.
Hydroxide Ion Concentration
The concentration of hydroxide ions \(\left[\mathrm{OH}^{-}\right]\) in a neutral solution is equal to that of hydrogen ions \(\left[\mathrm{H}^{+}\right]\).This balance is crucial for maintaining neutrality within a solution.When \(\left[\mathrm{H}^{+}\right]\) is known, \(\left[\mathrm{OH}^{-}\right]\) can be directly determined in neutral solutions. For example, at 0°C where \(K_w = 1.2 \times 10^{-15}\), if \(\left[\mathrm{H}^{+}\right] = 3.5 \times 10^{-8}\) M:
Then, by definition of neutrality:\[\left[\mathrm{OH}^{-}\right] = \left[\mathrm{H}^{+}\right]\]Thus, \(\left[\mathrm{OH}^{-}\right] \approx 3.5 \times 10^{-8} \) M.This equality simplifies calculations, allowing us to focus on either ion and infer the other's concentration instantly.It highlights how neutrality in solutions is defined not by the exact value of ion concentrations but by their equality at a given temperature.Such understanding assists with predicting solution behavior as it transitions between neutral, acidic, or basic states.

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Most popular questions from this chapter

Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) causing the pH of clean, unpolluted rain to range from about 5.2 to 5.6. What are the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops?

Indicate whether each of the following statements is correct or incorrect. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}(a q) \rightleftharpoons\)

Phenylacetic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\right)\) is one of the substances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A 0.085\(M\) solution of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\) has a pH of \(2.68 .\) Calculate the \(K_{a}\) value for this acid.

Predict which member of each pair produces the more acidic aqueous solution: (a) \(\mathrm{K}^{+}\) or \(\mathrm{Cu}^{2+},(\mathbf{b}) \mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) , (c) \(\mathrm{Al}^{3+}\) or \(\mathrm{Ga}^{3+}\) .

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