Question

Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution at \(25^{\circ} \mathrm{C}\). Identify each solution as neutral, acidic, or basic. a. \(\mathrm{pH}=7.40\) (the normal \(\mathrm{pH}\) of blood) b. \(\mathrm{pH}=15.3\) c. \(\mathrm{pH}=-1.0\) d. \(p H=3.20\) e. \(\mathrm{pOH}=5.0\) f. \(p O H=9.60\)

Short answer

#tag_title#d. Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for pH=3.20 #tag_content#Use the same formulas as before. Calculate \(\left[\mathrm{H}^{+}\right]\) for pH=3.20: \( \left[\mathrm{H}^{+}\right] = 10^{-3.20} \) \( \left[\mathrm{H}^{+}\right] = 6.31\times10^{-4} \textnormal{M} \) Now calculate pOH and \(\left[\mathrm{OH}^{-}\right]\): \( pOH = 14 - 3.20 = 10.80 \) \( \left[\mathrm{OH}^{-}\right] = 10^{-10.80} \) \( \left[\mathrm{OH}^{-}\right] = 1.58\times10^{-11} \textnormal{M} \) Since \(\left[\mathrm{H}^{+}\right] (6.31\times10^{-4}) > \left[\mathrm{OH}^{-}\right] (1.58\times10^{-11})\); the solution is acidic. #tag_title#e. Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for pOH=5.0 #tag_content#Use the formulas for pOH and \(\left[\mathrm{OH}^{-}\right]\) relationship: \( \left[\mathrm{OH}^{-}\right] = 10^{-pOH} \) \( \left[\mathrm{OH}^{-}\right] = 10^{-5} \) \( \left[\mathrm{OH}^{-}\right] = 1\times10^{-5} \textnormal{M} \) Now calculate pH and \(\left[\mathrm{H}^{+}\right]\): \( pH = 14 - 5 = 9 \) \( \left[\mathrm{H}^{+}\right] = 10^{-9} \) \( \left[\mathrm{H}^{+}\right] = 1\times10^{-9} \textnormal{M} \) Since \(\left[\mathrm{H}^{+}\right] (1\times10^{-9}) < \left[\mathrm{OH}^{-}\right] (1\times10^{-5})\); the solution is basic. #tag_title#f. Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for pOH=9.60 #tag_content#Calculate \(\left[\mathrm{OH}^{-}\right]\) for pOH=9.60: \( \left[\mathrm{OH}^{-}\right] = 10^{-9.60} \) \( \left[\mathrm{OH}^{-}\right] = 2.51\times10^{-10} \textnormal{M} \) Now calculate pH and \(\left[\mathrm{H}^{+}\right]\): \( pH = 14 - 9.60 = 4.40 \) \( \left[\mathrm{H}^{+}\right] = 10^{-4.40} \) \( \left[\mathrm{H}^{+}\right] = 3.98\times10^{-5} \textnormal{M} \) Since \(\left[\mathrm{H}^{+}\right] (3.98\times10^{-5}) > \left[\mathrm{OH}^{-}\right] (2.51\times10^{-10})\); the solution is acidic.

Step-by-step solution

a. Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for pH=7.4

Use the formula for pH and \(\left[\mathrm{H}^{+}\right]\) relationship: \( pH = -\log\left[\mathrm{H}^{+}\right] \) Rearrange the formula to solve for \(\left[\mathrm{H}^{+}\right]\): \( \left[\mathrm{H}^{+}\right] = 10^{-pH} \) Now, plug the given pH value (7.4) into the formula: \( \left[\mathrm{H}^{+}\right] = 10^{-7.4} \) Calculate the \(\left[\mathrm{H}^{+}\right]\): \( \left[\mathrm{H}^{+}\right] = 3.98\times10^{-8} \textnormal{M} \) Now let's calculate the \(\left[\mathrm{OH}^{-}\right]\) by using the pH-pOH relationship: \( pOH = 14 - pH \) \( pOH = 14 - 7.4 = 6.6 \) Now, use the pOH and \(\left[\mathrm{OH}^{-}\right]\) relationship: \( pOH = -\log\left[\mathrm{OH}^{-}\right] \) \( \left[\mathrm{OH}^{-}\right] = 10^{-pOH} \) Plug the calculated pOH value (6.6) into the formula: \( \left[\mathrm{OH}^{-}\right] = 10^{-6.6} \) Calculate the \(\left[\mathrm{OH}^{-}\right]\): \( \left[\mathrm{OH}^{-}\right] = 2.51\times10^{-7} \textnormal{M} \) Since \(\left[\mathrm{H}^{+}\right] \neq \left[\mathrm{OH}^{-}\right]\) and \(\left[\mathrm{H}^{+}\right] (3.98\times10^{-8}) < \left[\mathrm{OH}^{-}\right] (2.51\times10^{-7})\); the solution is basic.

b. Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for pH=15.3

We will use the same formulas as in part a. First, calculate \(\left[\mathrm{H}^{+}\right]\) for pH=15.3: \( \left[\mathrm{H}^{+}\right] = 10^{-15.3} \) \( \left[\mathrm{H}^{+}\right] = 5.01\times10^{-16} \textnormal{M} \) Now calculate pOH and \(\left[\mathrm{OH}^{-}\right]\): \( pOH = 14 - 15.3 = -1.3 \) \( \left[\mathrm{OH}^{-}\right] = 10^{-(-1.3)} \) \( \left[\mathrm{OH}^{-}\right] = 19.95 \textnormal{M} \) Since \(\left[\mathrm{H}^{+}\right] (5.01\times10^{-16}) < \left[\mathrm{OH}^{-}\right] (19.95)\); the solution is basic.

c. Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for pH=-1

Again, use the same formulas used before. Calculate \(\left[\mathrm{H}^{+}\right]\) for pH=-1: \( \left[\mathrm{H}^{+}\right] = 10^{1} \) \( \left[\mathrm{H}^{+}\right] = 10 \textnormal{M} \) Now calculate pOH and \(\left[\mathrm{OH}^{-}\right]\): \( pOH = 14 - (-1) = 15 \) \( \left[\mathrm{OH}^{-}\right] = 10^{-15} \) \( \left[\mathrm{OH}^{-}\right] = 1\times10^{-15} \textnormal{M} \) Since \(\left[\mathrm{H}^{+}\right] (10) > \left[\mathrm{OH}^{-}\right] (1\times10^{-15})\); the solution is acidic.

Key concepts

Acidic Solutions

Acidic solutions are chemical solutions with a high concentration of hydrogen ions, also known as hydronium ions, when dissolved in water. This abundance of hydronium ions (\( \text{H}^+ \))gives acidic solutions a pH less than 7.
When we talk about solutions like lemon juice or vinegar, we are referring to everyday acidic solutions. These solutions taste sour and can be corrosive.
For example, in the original problem, a solution with a pH of -1 is extremely acidic, which means it holds a very high concentration of hydronium ions, \( \left[ \text{H}^+ \right] = 10 \text{M} \).

• The more the hydronium ions, the lower the pH number.
• Acidic solutions have more \( \text{H}^+ \) ions than \( \text{OH}^- \) ions.

Understanding the properties of acidic solutions is crucial in many fields such as chemistry, biology, and even in everyday life.
For instance, many biochemical processes within our bodies rely on the controlled presence of acidic conditions.

Basic Solutions

Basic solutions, also known as alkaline solutions, have a higher concentration of hydroxide ions, (\( \text{OH}^- \)), than hydrogen ions, giving them a pH greater than 7. A common example of a basic solution is toothpaste.

• Basic solutions often feel slippery and can taste bitter.
• They commonly include substances such as baking soda or soap.

In the original exercise, a solution with a pH of 15.3 is described as basic, which indicates a very low concentration of hydrogen ions and a high concentration of hydroxide ions.
If you calculate the \( \left[ \text{OH}^- \right] \) using the \( pOH \) formula (\( \text{pOH} = 14 - pH \)),you find this solution is indeed strongly basic with a hydroxide ion concentration of \( 19.95 \text{M} \).Understanding the nature of basic solutions is vital in sectors like cleaning products manufacturing and food preparation, as they help in neutralizing acidic substances.

Neutral Solutions

Neutral solutions have a balanced concentration of both hydronium ions, (\( \text{H}^+ \)), and hydroxide ions, (\( \text{OH}^- \)). These solutions typically have a pH of 7. The most classic example of a neutral solution is pure water. Neutral solutions are neither acidic nor basic, making them indispensable in many scientific and everyday applications.
If you consider household items, water is the most common neutral solution, though, in reality, pure water is rarely found outside laboratory conditions.

• Neutral pH is important in biological systems where extreme pH could damage living cells.
• In the exercise, solutions with a pH near 7, like blood with a pH of 7.4, are often slightly basic but close to neutral.

Understanding neutral solutions forms the foundation for topics in chemistry and biology, where the balance between hydronium and hydroxide ions is vital for maintaining life.

Hydronium Ion Concentration

Hydronium ion concentration, denoted as (\( \left[ \text{H}^+ \right] \)), is a critical measure that indicates how acidic a solution is.
It stems from the ionization of an acid in water. The hydronium ion concentration is directly used to calculate the pH of a solution using the formula: \( pH = -\log_{10} \left(\left[ \text{H}^+ \right]\right) \). A higher concentration of these ions results in a lower pH value, signifying a stronger acid.

• In very acidic solutions, hydronium ion concentration is high, causing a low pH.
• In the original exercise, the different pH values help us calculate the hydronium concentration using the given formula.

For example, for pH 3.20, the hydronium ion concentration would be \( \left[ \text{H}^+ \right] = 10^{-3.20} \),giving a tangible measure of acids like vinegar, which might have a pH around this value.
Understanding hydronium ion concentration helps in fields like environmental science for assessing soil acidity and water quality.

Hydroxide Ion Concentration

Hydroxide ion concentration, (\(\left[ \text{OH}^- \right] \)), plays a significant role in determining the basicity of a solution.
It results from the dissociation of a base in water. The \(\left[ \text{OH}^- \right]\) can be calculated using \( pOH = -\log_{10} \left(\left[ \text{OH}^- \right]\right) \).
This approach helps in identifying solutions which are basic.

• A high hydroxide ion concentration usually results in a high pH, indicating a basic solution.
• In the exercise, the pH or pOH values are used to find out the \(\left[ \text{OH}^- \right]\) of solutions.

For instance, a solution with a pOH of 5.0 can be calculated for hydroxide concentration.The hydroxide ions in basic solutions help in many processes like neutralizing acids in antacids.
Understanding \(\left[ \text{OH}^- \right]\) is essential in many sectors, such as wastewater treatment, where it helps maintain optimal pH levels for fish and aquatic plants.