2. Suppose you are given a strong electrolyte [tex]HCl[/tex] of [tex]0.5 \, M[/tex] concentration and a weak electrolyte [tex]NH_4OH[/tex] of [tex]0.5 \, M[/tex] concentration. Given the [tex]\overline{Kb}[/tex] of [tex]NH_4OH = 1.8 \times 10^{-5}[/tex] and [tex]Ka[/tex] for [tex]HCl = 1.3 \times 10^6[/tex].

a. Calculate the [tex]pH[/tex] of [tex]0.5 \, M \, HCl[/tex] and [tex]0.5 \, M \, NH_4OH[/tex].



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.