Answer :
Sure, let's solve the given problem step by step.
### Part A: Calculate the pH of 0.5 M HCl
1. Identify the given data:
- Concentration of HCl = 0.5 M
- Since HCl is a strong acid, it dissociates completely in water.
2. Calculate the [tex]\([H^+]\)[/tex] concentration of HCl:
- For a strong acid like HCl, the [tex]\([H^+]\)[/tex] concentration is equal to the concentration of the acid.
- [tex]\([H^+] = 0.5\)[/tex] M
3. Calculate the pH:
- The pH is calculated using the formula [tex]\( \text{pH} = -\log_{10}[\text{H}^+] \)[/tex]
- Substituting the values, [tex]\(\text{pH} = -\log_{10}(0.5)\)[/tex]
4. Result:
- The pH of 0.5 M HCl is approximately [tex]\(0.30\)[/tex]
### Part B: Calculate the pH of 0.5 M NH4OH
1. Identify the given data:
- Concentration of NH4OH = 0.5 M
- [tex]\(Kb\)[/tex] of NH4OH = [tex]\(1.8 \times 10^{-5}\)[/tex]
- NH4OH is a weak base, so it does not dissociate completely in water.
2. Calculate the [tex]\([OH^-]\)[/tex] concentration of NH4OH:
- For a weak base, [tex]\([OH^-]\)[/tex] concentration is given by:
[tex]\[ \text{[OH}^-] = \sqrt{K_b \times \text{concentration of the base}} \][/tex]
- Substituting the values:
[tex]\[ \text{[OH}^-] = \sqrt{1.8 \times 10^{-5} \times 0.5} \][/tex]
- Simplifying, [tex]\(\text{[OH}^-] \approx 0.003\)[/tex] M
3. Calculate the pOH:
- The pOH is calculated using the formula [tex]\( \text{pOH} = -\log_{10}[\text{OH}^-] \)[/tex]
- Substituting the values, [tex]\(\text{pOH} = -\log_{10}(0.003)\)[/tex]
4. Calculate the pH:
- We know that [tex]\( \text{pH} + \text{pOH} = 14 \)[/tex]
- Therefore, [tex]\( \text{pH} = 14 - \text{pOH} \)[/tex]
- Substituting the pOH value, [tex]\(\text{pH} = 14 - 2.52\)[/tex]
5. Result:
- The pOH is approximately [tex]\(2.52\)[/tex]
- The pH of 0.5 M NH4OH is approximately [tex]\(11.48\)[/tex]
### Summary
- pH of 0.5 M HCl: [tex]\(0.30\)[/tex]
- pH of 0.5 M NH4OH: [tex]\(11.48\)[/tex]
These are the calculated pH values for the given solutions.
### Part A: Calculate the pH of 0.5 M HCl
1. Identify the given data:
- Concentration of HCl = 0.5 M
- Since HCl is a strong acid, it dissociates completely in water.
2. Calculate the [tex]\([H^+]\)[/tex] concentration of HCl:
- For a strong acid like HCl, the [tex]\([H^+]\)[/tex] concentration is equal to the concentration of the acid.
- [tex]\([H^+] = 0.5\)[/tex] M
3. Calculate the pH:
- The pH is calculated using the formula [tex]\( \text{pH} = -\log_{10}[\text{H}^+] \)[/tex]
- Substituting the values, [tex]\(\text{pH} = -\log_{10}(0.5)\)[/tex]
4. Result:
- The pH of 0.5 M HCl is approximately [tex]\(0.30\)[/tex]
### Part B: Calculate the pH of 0.5 M NH4OH
1. Identify the given data:
- Concentration of NH4OH = 0.5 M
- [tex]\(Kb\)[/tex] of NH4OH = [tex]\(1.8 \times 10^{-5}\)[/tex]
- NH4OH is a weak base, so it does not dissociate completely in water.
2. Calculate the [tex]\([OH^-]\)[/tex] concentration of NH4OH:
- For a weak base, [tex]\([OH^-]\)[/tex] concentration is given by:
[tex]\[ \text{[OH}^-] = \sqrt{K_b \times \text{concentration of the base}} \][/tex]
- Substituting the values:
[tex]\[ \text{[OH}^-] = \sqrt{1.8 \times 10^{-5} \times 0.5} \][/tex]
- Simplifying, [tex]\(\text{[OH}^-] \approx 0.003\)[/tex] M
3. Calculate the pOH:
- The pOH is calculated using the formula [tex]\( \text{pOH} = -\log_{10}[\text{OH}^-] \)[/tex]
- Substituting the values, [tex]\(\text{pOH} = -\log_{10}(0.003)\)[/tex]
4. Calculate the pH:
- We know that [tex]\( \text{pH} + \text{pOH} = 14 \)[/tex]
- Therefore, [tex]\( \text{pH} = 14 - \text{pOH} \)[/tex]
- Substituting the pOH value, [tex]\(\text{pH} = 14 - 2.52\)[/tex]
5. Result:
- The pOH is approximately [tex]\(2.52\)[/tex]
- The pH of 0.5 M NH4OH is approximately [tex]\(11.48\)[/tex]
### Summary
- pH of 0.5 M HCl: [tex]\(0.30\)[/tex]
- pH of 0.5 M NH4OH: [tex]\(11.48\)[/tex]
These are the calculated pH values for the given solutions.
