Question

Determine the pH of each solution. a. \(1.0 \times 10^{-2} \mathrm{M} \mathrm{NaOH}\) b. \(1.0 \times 10^{-3} \mathrm{M} \mathrm{M} \mathrm{KOH}\) c. \(1.0 \times 10^{-4} \mathrm{M} \mathrm{LiOH}\)

Short answer

a. 12, b. 11, c. 10

Step-by-step solution

Understanding pH and pOH

pH is a measure of the acidity of a solution, while pOH measures the basicity. The relationship between pH and pOH is given by the equation: \[\text{pH} + \text{pOH} = 14\]

Calculate pOH for each solution

For a solution of a strong base like NaOH, KOH, or LiOH, the concentration of OH⁻ ions is equal to the concentration of the base. Use the formula: \[\text{pOH} = -\text{log}[\text{OH}^-]\]

Calculate pH from pOH

Using the relationship: \[\text{pH} + \text{pOH} = 14\], \ \text{pH} can be derived by rearranging the equation: \[\text{pH} = 14 - \text{pOH}\]

Solve for the first solution

Given: \[\text{Concentration of NaOH} = 1.0 \times 10^{-2} \text{ M}\].Calculate pOH: \[\text{pOH} = - \text{log}(1.0 \times 10^{-2}) = 2\]Then, calculate pH: \[\text{pH} = 14 - 2 = 12\]

Solve for the second solution

Given: \[\text{Concentration of KOH} = 1.0 \times 10^{-3} \text{ M}\].Calculate pOH: \[\text{pOH} = - \text{log}(1.0 \times 10^{-3}) = 3\]Then, calculate pH: \[\text{pH} = 14 - 3 = 11\]

Solve for the third solution

Given: \[\text{Concentration of LiOH} = 1.0 \times 10^{-4} \text{ M}\].Calculate pOH: \[\text{pOH} = - \text{log}(1.0 \times 10^{-4}) = 4\]Then, calculate pH: \[\text{pH} = 14 - 4 = 10\]

Key concepts

pOH calculation

To calculate pOH, you need to understand that it is directly related to the concentration of hydroxide ions \( \text{OH}^- \) in the solution.
In solutions with strong bases like NaOH, KOH, and LiOH, the concentration of \( \text{OH}^- \text{ ions} \) equals the concentration of the base itself.
The formula you will use to find pOH is: \[ \text{pOH} = -\text{log}[\text{OH}^-] \]
Take note that the \( \text{log} \) function here refers to the common logarithm (base 10).
For example, to find the pOH of a \( 1.0 \times 10^{-2} \text{ M} \text{ NaOH} \) solution, you perform the calculation:\[ \text{pOH} = -\text{log}(1.0 \times 10^{-2}) = 2 \]
This simple logarithmic calculation translates to understanding the acidity or basicity of the solution.

strong bases

Strong bases are substances that completely dissociate in water, meaning they break apart into ions completely.
Examples include NaOH (sodium hydroxide), KOH (potassium hydroxide), and LiOH (lithium hydroxide).
When these compounds dissolve, they produce hydroxide ions \( (\text{OH}^-) \), contributing to the solution's basicity.
For instance, if you have \( 1.0 \times 10^{-2} \text{ M NaOH} \), it dissociates fully to give \( 1.0 \times 10^{-2} \text{ M OH}^- \).
This complete dissociation is a key factor when calculating pH and pOH for solutions with strong bases.
Knowing the concentration of the base allows you to directly find the concentration of \( \text{OH}^- \) ions, simplifying the calculation process.

logarithmic functions in chemistry

Logarithmic functions play a crucial role in chemistry, especially in calculating pH and pOH.
The common logarithm (log base 10) is used to convert concentrations of ions into a more manageable scale.
The logarithm of a number tells you the power to which 10 must be raised to get that number.
For example, \( \text{log}(10^{-2}) = -2 \).
When calculating pOH, this concept helps compress large ranges of values into a smaller, understandable scale.
Using the formula \[ \text{pOH} = -\text{log}[\text{OH}^-] \], you apply negative logarithms to express concentrations of \( \text{OH}^- \) ions in a simpler form.
This makes interpreting and comparing the acidity or basicity of different solutions straightforward.

acid-base equilibrium

Acid-base equilibrium refers to the balance between acidic and basic species in a solution.
In any aqueous solution, water naturally dissociates into hydrogen ions \( (\text{H}^+) \) and hydroxide ions \( (\text{OH}^-) \).
The product of their concentrations is constant at a given temperature and expressed as \[ \text{Kw} = [\text{H}^+][\text{OH}^-] \].
At 25°C, \( \text{Kw} \) is \( 1.0 \times 10^{-14} \).
Using this relation, you derive the equation connecting pH and pOH: \[ \text{pH} + \text{pOH} = 14 \].
This shows that if you know the pOH of a solution, you can always find the pH by subtracting pOH from 14.
Balancing these values helps understand whether a solution is acidic or basic and to what extent.
For strong bases, which dissociate completely, calculating pH from pOH becomes a straightforward task using the equilibrium relationship.