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Calculate the \(\mathrm{pH}\) of each solution. (a) \(\left[\mathrm{OH}^{-}\right]=2.8 \times 10^{-11} \mathrm{M}\) (b) \(\left[\mathrm{OH}^{-}\right]=9.6 \times 10^{-3} \mathrm{M}\) (c) \(\left[\mathrm{OH}^{-}\right]=3.8 \times 10^{-12} \mathrm{M}\) (d) \(\left[\mathrm{OH}^{-}\right]=6.4 \times 10^{-4} \mathrm{M}\)

Short Answer

Expert verified
The pH of each solution: (a) ~10.55, (b) ~11.02, (c) ~10.42, (d) ~11.19

Step by step solution

01

- Understanding pH and pOH

The pH of a solution is a measure of its acidity or alkalinity. The pH and pOH of a solution are related by the equation \( pH + pOH = 14 \), where 14 is the sum at 25 degrees Celsius. The pOH can be calculated using the hydroxide ion concentration \( [OH^{-}] \) with the formula \( pOH = -\log[OH^{-}] \) for any given solution.
02

- Calculate pOH for (a)

First, calculate the pOH for the given concentration of hydroxide ions. For \( [OH^{-}] = 2.8 \times 10^{-11} \mathrm{M} \), calculate \( pOH = -\log(2.8 \times 10^{-11}) \) using a calculator.
03

- Calculate pH for (a)

Having calculated the pOH, use the relationship \( pH + pOH = 14 \) to find the pH. Subtract the pOH from 14 to get the pH.
04

- Calculate pOH for (b)

Repeat the same process for \( [OH^{-}] = 9.6 \times 10^{-3} \mathrm{M} \) to find the pOH. Use the formula \( pOH = -\log(9.6 \times 10^{-3}) \).
05

- Calculate pH for (b)

Again, use the relationship \( pH + pOH = 14 \) to find the pH by subtracting the previously calculated pOH from 14.
06

- Calculate pOH for (c)

For \( [OH^{-}] = 3.8 \times 10^{-12} \mathrm{M} \) calculate the pOH using \( pOH = -\log(3.8 \times 10^{-12}) \).
07

- Calculate pH for (c)

Determine the pH from the calculated pOH using the relationship \( pH + pOH = 14 \) and subtracting the pOH from 14.
08

- Calculate pOH for (d)

Similarly, for \( [OH^{-}] = 6.4 \times 10^{-4} \mathrm{M} \) calculate the pOH using \( pOH = -\log(6.4 \times 10^{-4}) \).
09

- Calculate pH for (d)

Finally, find the pH for part (d) by subtracting the calculated pOH from 14, using the relationship \( pH + pOH = 14 \) once again.

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Key Concepts

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

pOH calculation
Understanding the pOH of a solution is crucial when examining its chemical properties, specifically its acidity and alkalinity. The pOH is an expression of the hydroxide ion concentration in a solution, and it's determined by taking the negative logarithm of the hydroxide ion concentration.

To calculate the pOH, use the formula:
\[ pOH = -\text{log}([OH^-]) \]
This formula requires a scientific calculator with a logarithm function, usually denoted as 'log'. For instance, if you have a hydroxide ion concentration of \( 2.8 \times 10^{-11} M \), you would input this into your calculator to find the pOH value.

The importance of calculating pOH lies in its relationship with pH, which together help us understand whether a solution is acidic or basic. Remember, a lower pOH indicates a higher concentration of hydroxide ions, which correlates with a more basic solution. Conversely, a higher pOH means fewer hydroxide ions, suggesting a more acidic solution.
Hydroxide Ion Concentration
The hydroxide ion concentration, denoted as \( [OH^-] \), measures the amount of hydroxide ions in a solution and is key to determining the solution's acidity or basicity. It's typically expressed in moles per liter (M), where one mole is Avogadro's number of hydroxide ions.

An understanding of hydroxide ion concentration is foundational because it directly affects the calculation of both pH and pOH. For instance, knowing that the hydroxide ion concentration in a solution is \( 9.6 \times 10^{-3} M \), we can infer that the solution has a significant amount of hydroxide ions, and is likely to be basic.

To put it in perspective, pure water at 25°C has a hydroxide ion concentration of \( 1 \times 10^{-7} M \), considered neutral. Any concentration above this level indicates basicity, while any concentration below it suggests acidity.
Acidity and Alkalinity
Acidity and alkalinity are terms used to describe the chemical character of a solution. Acidity refers to the presence of hydrogen ions (H+) or a lower pH, while alkalinity refers to the presence of hydroxide ions (OH-) or a higher pH. Pure water is neutral, with a pH of 7, but when a substance is dissolved in water it can either increase the hydrogen ion concentration, making the solution acidic, or increase the hydroxide ion concentration, making the solution basic or alkaline.

pH Scale

The pH scale ranges from 0 to 14, with 7 being neutral. A pH below 7 indicates an acidic solution, while a pH above 7 indicates a basic (alkaline) solution. This scale is logarithmic, which means that each whole number on the scale represents a tenfold increase or decrease in acidity or alkalinity. For example, a solution with a pH of 4 is ten times more acidic than a solution with a pH of 5.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each solution. (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.7 \times 10^{-8} \mathrm{M}\) (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.0 \times 10^{-7} \mathrm{M}\) (c) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.2 \times 10^{-6} \mathrm{M}\) (d) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.4 \times 10^{-4} \mathrm{M}\)

Give a possible value of \(\left[\mathrm{OH}^{-}\right]\)and \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)in a solution that is: (a) acidic (b) basic (c) neutral

Write the formula for the conjugate acid of each base. (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (c) \(\mathrm{Cl}^{-}\) (d) \(\mathrm{F}^{-}\)

Calculate \(\left[\mathrm{OH}^{-}\right]\)given \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)in each aqueous solution and classify the solution as acidic or basic. (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.3 \times 10^{-3} \mathrm{M}\) (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.1 \times 10^{-12} \mathrm{M}\) (c) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5.2 \times 10^{-4} \mathrm{M}\) (d) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=6.1 \times 10^{-9} \mathrm{M}\)

Determine \(\left[\mathrm{OH}^{-}\right]\)in each base solution. If the acid is weak, indicate the value that \(\left[\mathrm{OH}^{-}\right]\)is less than. (a) \(2.5 \mathrm{M} \mathrm{KOH}\) (b) \(1.95 \mathrm{M} \mathrm{NH}_{3}\) (c) \(0.225 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) (d) \(1.8 \mathrm{MC}_{5} \mathrm{H}_{5} \mathrm{~N}\)

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