Problem 106
Question
What is the pH of a solution that is \(1.2 \times 10^{-8} \mathrm{M}\) in \(\mathrm{KOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?
Step-by-Step Solution
Verified Answer
The pH of the solution containing \(1.2 \times 10^{-8} \:M \) KOH is 6.08. This answer seems unusual because the pH is less than 7, indicating slight acidity rather than basicity. The issue arises from ignoring the autodissociation of water, which is not a valid assumption in this case due to the extremely low concentration of KOH.
1Step 1: Calculate the concentration of hydroxide ions
Since KOH is a strong base and dissociates fully in water, the concentration of hydroxide ions (OH-) is equal to the concentration of KOH in the solution. In this case, the concentration of hydroxide ions is \(1.2 \times 10^{-8} \:M \).
2Step 2: Calculate the pOH
Using the formula for pOH, we have:
\[ pOH = -\log_{10} [\mathrm{OH^-}] \]
So, we get:
\[ pOH = -\log_{10} (1.2 \times 10^{-8}) = 7.92 \]
3Step 3: Calculate the pH
The relationship between pH and pOH is given by the equation:
\[ pH + pOH = 14 \]
Using the pOH calculated in Step 2, we can now find the pH:
\[ pH = 14 - pOH = 14 - 7.92 = 6.08 \]
The pH of the solution containing KOH with a concentration of \(1.2 \times 10^{-8} \:M\) is 6.08.
4Step 4: Check the validity of the answer and assumptions
Our calculated pH value seems a bit odd because it is smaller than 7, which is unusual for a basic solution. The explanation for this is that the slight acidity of the solution comes from the autodissociation of water. The reason why this error occurred is because we assumed that the amount of OH- ions came exclusively from the dissociation of KOH, while ignoring the autodissociation of water.
Normally, the assumption is valid when dealing with strong acids or bases with relatively high concentrations, as the contribution of water's autodissociation becomes less significant. However, in this case, the concentration of KOH is extremely low, which makes the autodissociation of water more important and renders our assumption invalid.
Key Concepts
Strong Base DissociationAutodissociation of WaterpOH Calculation
Strong Base Dissociation
Strong base dissociation is a fundamental concept to grasp when calculating pH in solutions involving bases like potassium hydroxide (KOH). A strong base dissociates completely in water. This means that each molecule of KOH breaks apart into potassium ions (
K^+
) and hydroxide ions (
OH^-
).
To determine the concentration of hydroxide ions, you rely directly on the initial concentration of the KOH solution. In the initial example, given a concentration of 1.2 imes 10^{-8} ext{M} in KOH , it directly translates to the same concentration of hydroxide ions because the dissociation is "complete."
To determine the concentration of hydroxide ions, you rely directly on the initial concentration of the KOH solution. In the initial example, given a concentration of 1.2 imes 10^{-8} ext{M} in KOH , it directly translates to the same concentration of hydroxide ions because the dissociation is "complete."
- The concept of complete dissociation is crucial for understanding how strong bases impact the pH or pOH of a solution.
- In other words, for strong bases, OH^- concentration equals the initial base concentration.
Autodissociation of Water
The autodissociation of water is another important chemical process impacting
pH
and
pOH
calculations, particularly in very dilute solutions. In pure water, molecules auto-dissociate into hydrogen ions (
H^+
) and hydroxide ions (
OH^-
), even in the absence of other substances. This dissociation constant for water is known as
K_w
and equals
1 imes 10^{-14}
at 25°C.
For very low concentrations of strong bases like 1.2 imes 10^{-8} ext{M} KOH, the autodissociation of water becomes significant. In low concentrations, the OH^- ions from the dissociation of KOH can nearly equal those from water's self-ionization, thus affecting the solution's overall pH .
For very low concentrations of strong bases like 1.2 imes 10^{-8} ext{M} KOH, the autodissociation of water becomes significant. In low concentrations, the OH^- ions from the dissociation of KOH can nearly equal those from water's self-ionization, thus affecting the solution's overall pH .
- This effect is sometimes overlooked in concentrated solutions, but it's essential to consider in dilute cases.
- At equilibrium, the relation pH + pOH = 14 holds, provided we factor in the ions from water itself.
pOH Calculation
The pOH of a solution is a measure of its hydroxide ion concentration. It's essential for determining the corresponding pH value, completing the entire picture of acidity or basicity in a given solution. The pOH is calculated using the formula: \[ pOH = -\log_{10} [\text{OH}^- \text{ concentration}] \]
In our given exercise, the hydroxide ion concentration found was 1.2 \times 10^{-8} ext{M}. Applying the pOH formula, we get: \[ pOH = -\log_{10} (1.2 \times 10^{-8}) \approx 7.92 \]
In our given exercise, the hydroxide ion concentration found was 1.2 \times 10^{-8} ext{M}. Applying the pOH formula, we get: \[ pOH = -\log_{10} (1.2 \times 10^{-8}) \approx 7.92 \]
- This value of pOH, when used in the equation pH + pOH = 14, will help determine the solution's pH.
- The calculation confirms knowledge of basic logarithmic and chemical principles when deriving the solution's basicity.
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