Sure, let's solve each question step-by-step.
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Question 4:
A certain acid has a pH of 4.5. What would be the complimentary pOH value?
Solution:
1. The relationship between pH and pOH can be expressed using the formula:
[tex]\[
\text{pH} + \text{pOH} = 14
\][/tex]
2. Given the pH of the acid is 4.5, we can substitute this value into the equation:
[tex]\[
4.5 + \text{pOH} = 14
\][/tex]
3. Solving for pOH:
[tex]\[
\text{pOH} = 14 - 4.5 = 9.5
\][/tex]
So, the complimentary pOH value is 9.5.
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Question 5:
What is the pH of 0.0235M HCl solution?
Solution:
1. Hydrochloric acid (HCl) is a strong acid, which means it completely dissociates in water. Thus, the concentration of H\(^+\) ions is equal to the concentration of the HCl solution, which is 0.0235M.
2. The pH is calculated using the formula:
[tex]\[
\text{pH} = -\log_{10}[\text{H}^+]
\][/tex]
3. Substituting the given concentration:
[tex]\[
\text{pH} = -\log_{10}(0.0235)
\][/tex]
4. The calculated pH value is approximately:
[tex]\[
\text{pH} \approx 1.6289321377282637
\][/tex]
So, the pH of the 0.0235M HCl solution is approximately 1.6289.
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Question 6:
What is the pH of a 6.50 x 10^-6 M KOH solution?
Solution:
1. Potassium hydroxide (KOH) is a strong base, which means it completely dissociates in water. Thus, the concentration of OH\(^-\) ions is equal to the concentration of the KOH solution, which is 6.50 x 10^-6 M.
2. The pOH is calculated using the formula:
[tex]\[
\text{pOH} = -\log_{10}[\text{OH}^-]
\][/tex]
3. Substituting the given concentration:
[tex]\[
\text{pOH} = -\log_{10}(6.50 \times 10^{-6})
\][/tex]
4. The calculated pOH value is approximately:
[tex]\[
\text{pOH} \approx 5.187086643357144
\][/tex]
5. The pH can be found using the relationship:
[tex]\[
\text{pH} + \text{pOH} = 14
\][/tex]
6. Substituting the calculated pOH value:
[tex]\[
\text{pH} = 14 - 5.187086643357144
\][/tex]
7. The calculated pH value is approximately:
[tex]\[
\text{pH} \approx 8.812913356642856
\][/tex]
So, the pH of the 6.50 x 10^-6 M KOH solution is approximately 8.8129.
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