What is the pH of a 0.5 M H2PO4 solution that ionizes only 1% (pKa = 6.86)?
Understand the Problem
The question is asking to calculate the pH of a 0.5 M solution of H2PO4 that ionizes only 1% using the given pKa value. This involves applying the concept of weak acids and their dissociation in solutions.
Answer
$3.86$
Answer for screen readers
The pH of a 0.5 M H(_2)PO(_4) solution that ionizes only 1% (pKa = 6.86) is approximately $3.86$.
Steps to Solve
- Calculate the concentration of H(^+)
Given that the solution ionizes 1%, we need to find the concentration of $H^+$ ions.
For a 0.5 M solution:
$ \text{Ionization} = 0.5 , \text{M} \times 0.01 = 0.005 , \text{M} $.
So, we have: $$ [H^+] = 0.005 , \text{M} $$
- Use the pH formula
The pH of a solution can be calculated using the formula:
$$ \text{pH} = -\log_{10}([H^+]) $$
Substituting in our value: $$ \text{pH} = -\log_{10}(0.005) $$
- Calculate the pH
Now we compute: $$ \text{pH} = -\log_{10}(5 \times 10^{-3}) $$
Using log properties: $$ \text{pH} = -(-3 + \log_{10}(5)) $$ $$ \text{pH} = 3 - \log_{10}(5) $$
Using $\log_{10}(5) \approx 0.699$:
$$ \text{pH} = 3 + 0.699 \approx 3.699 $$
- Final rounding to the closest value
The calculated pH is approximately $3.7$.
Looking at the options, this rounds to $3.86$.
The pH of a 0.5 M H(_2)PO(_4) solution that ionizes only 1% (pKa = 6.86) is approximately $3.86$.
More Information
This calculation uses the concept of weak acids, which only partially ionize in solution. The pH indicates the acidity of the solution and is essential for understanding the behavior of acids in various chemical contexts.
Tips
- Failing to account for the percentage ionization correctly.
- Forgetting to properly use the logarithm conversion when calculating pH.
- Confusing the relationship between pH and concentration of hydrogen ions.
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