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By what factor does \(\left[\mathrm{H}^{+}\right]\) change for a pH change of (a) 2.00 units, (b) 0.50 units?

Short Answer

Expert verified
For a pH change of (a) 2.00 units, the factor by which \(\mathrm{[H^+]}\) changes is \(10^2\) or 100 times. For a pH change of (b) 0.50 units, the factor by which \(\mathrm{[H^+]}\) changes is \(10^{0.5}\) or approximately 3.16 times.

Step by step solution

01

Recall the definition of pH

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen ion concentration: \(pH = -\log_{10}{\mathrm{[H^+]}}\)
02

Write the pH change equation

Given a change in pH, we want to find the corresponding change in the hydrogen ion concentration. So if the pH changes by \(x\) units, we can write the equation as: \(\Delta pH = pH_2 - pH_1 = x\)
03

Calculate the Factor of Change

To find the factor by which \(\mathrm{[H^+]}\) changes, we first need to express the hydrogen ion concentrations for the two different pH values as powers of 10. By doing so, we'll be able to establish the factor of change between them.
04

Step 3(a): Calculate the factor of change for 2.00 units of pH change

Given \(\Delta pH = 2.00\), let's find the hydrogen ion concentration at the two pH values: \(pH_1 = -\log_{10}{\mathrm{[H_1^{+}]}} \) \(pH_2 = -\log_{10}{\mathrm{[H_2^{+}]}} \) Since pH decreases with increasing \(\mathrm{[H^+]}\), we can assume that \(pH_2 = pH_1 - 2.00\). Now, \(\mathrm{[H_2^{+}]} = 10^{-(pH_1 - 2)} = 10^{-(pH_1)} \times 10^2 = \mathrm{[H_1^{+}]} \times 10^2\) So, the change in hydrogen ion concentration is a factor of \(10^2\) or 100 times for a pH change of 2.00 units.
05

Step 3(b): Calculate the factor of change for 0.50 units of pH change

Given \(\Delta pH = 0.50\), let's find the hydrogen ion concentration at the two pH values: \(pH_1 = -\log_{10}{\mathrm{[H_1^{+}]}} \) \(pH_2 = -\log_{10}{\mathrm{[H_2^{+}]}} \) Since pH decreases with increasing \(\mathrm{[H^+]}\), we can assume that \(pH_2 = pH_1 - 0.50\). Now, \(\mathrm{[H_2^{+}]} = 10^{-(pH_1 - 0.5)} = 10^{-(pH_1)} \times 10^{0.5} = \mathrm{[H_1^{+}]} \times 10^{0.5}\) So, the change in hydrogen ion concentration is a factor of \(10^{0.5}\) or approximately 3.16 times for a pH change of 0.50 units.

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Key Concepts

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

Hydrogen Ion Concentration
Hydrogen ion concentration, often represented as \(\left[\text{H}^+\right]\), is a key parameter in understanding the acidity of solutions. It specifically tells us how many hydrogen ions are present per unit volume of a solution. A high concentration of hydrogen ions typically means a more acidic solution. The concentration is inversely related to pH, which means as \(\left[\text{H}^+\right]\) concentration increases, the pH decreases, indicating a stronger acid. Conversely, as the concentration of hydrogen ions decreases, the pH increases, suggesting the solution is moving towards being neutral or even basic.In practical terms, knowing \(\left[\text{H}^+\right]\) allows chemists and biologists to understand how substances will react in different environments. This is vital in processes such as drug delivery in medicine, where the pH of the stomach might affect how a medication dissolves and is absorbed.
Logarithms in Chemistry
Logarithms are an essential mathematical tool in chemistry due to their relationship with pH calculations and concentration scales. The pH scale is logarithmic, meaning each whole number change on the scalerelates to a tenfold change in the concentration of \(\left[\text{H}^+\right]\) ions.This is expressed in the equation: \( pH = -\log_{10}\left[\text{H}^+\right] \). With this expression, small changes in hydrogen ion concentration result in substantial changes on the pH scale. That's why a decrease in pH by 1 unit indicates that the hydrogen ion concentration has increased by a factor of 10. Using logarithms allows us to work with manageable numbers instead of countless decimal places, facilitating easier calculations and better comprehension of chemical reactions and properties.Understanding logarithms in chemistry helps interpret results better andallows scientists to predict how substances react in different conditions, critical in everything from industrial processes to environmental science.
Acid-Base Reactions
Acid-base reactions are fundamental chemical reactions in which an acid donates a hydrogen ion \(\text{H}^+\) to a base. This concept is crucial in understanding how substances interact and change during reactions.In an acid-base reaction, the pH can change significantly based on the concentrations involved.For instance, the addition of an acid to a solution increases the \(\left[\text{H}^+\right]\), lowering the pH and making the solution more acidic.Similarly, adding a base decreases \(\left[\text{H}^+\right]\), raising the pH, thus making the solution more basic or alkaline.The pH value is particularly importan in fields such as agriculture, where it impacts soil fertility and crop growth,and in medicine, where it can affect how well drugs work in the body. Acid-base reactions and pH changes are not just confined to laboratory settings. They are also critical in natural processes, such as the formation of limestone caves from the reaction of \(\text{H}_2\text{CO}_3\), a weak acid, with limestone, a base.

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Most popular questions from this chapter

(a) The hydrogen oxalate ion \(\left(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right)\) is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conjugate acid of \(\mathrm{HC}_{2} \mathrm{O}_{4}\) ? What is its conjugate base?

Deuterium oxide \(\left(\mathrm{D}_{2} \mathrm{O},\right.\) where \(\mathrm{D}\) is deuterium, the hydrogen- 2 isotope) has an ion-product constant, \(K_{w^{+}}\) of \(8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\). Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) for pure (neutral) \(\mathrm{D}_{2} \mathrm{O}\) at this temperature.

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{p} K_{a}\) of butyric acid is \(4.84 .\) (a) Calculate the \(\mathrm{p} K_{b}\) for the butyrate ion. (b) Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of butyric acid. (c) Calculate the pH of a \(0.050 \mathrm{M}\) solution of sodium butyrate.

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

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 \mathrm{M} ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;(\mathrm{c})\) a so- lution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).

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