Certainly! Let's solve the problem step by step.
### Step 1: Understand the Given Data
We are given:
- The concentration of the weak base ([tex]\( [\text{Base}] \)[/tex]) = 0.18 M
- The base dissociation constant ([tex]\( K_b \)[/tex]) = [tex]\( 4.2 \times 10^{-10} \)[/tex]
### Step 2: Calculate the Concentration of [tex]\( \text{OH}^- \)[/tex] Ions
For a weak base, we use the formula for the base dissociation equilibrium:
[tex]\[ \text{Base} + \text{H}_2\text{O} \rightleftharpoons \text{OH}^- + \text{BH}^+ \][/tex]
The concentration of hydroxide ions [tex]\( [\text{OH}^-] \)[/tex] is found using:
[tex]\[ [\text{OH}^-] = \sqrt{K_b \times [\text{Base}]} \][/tex]
Substitute the given values into the formula:
[tex]\[ [\text{OH}^-] = \sqrt{4.2 \times 10^{-10} \times 0.18} \][/tex]
After performing the calculation, we get:
[tex]\[ [\text{OH}^-] \approx 8.694826047713664 \times 10^{-6} \, \text{M} \][/tex]
### Step 3: Calculate the Concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex] Ions
The relationship between [tex]\( [\text{H}_3\text{O}^+] \)[/tex] and [tex]\( [\text{OH}^-] \)[/tex] is given by the water dissociation constant ([tex]\( K_w \)[/tex]) at 25°C:
[tex]\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \][/tex]
where [tex]\( K_w = 1.0 \times 10^{-14} \)[/tex].
Rearrange the formula to find [tex]\( [\text{H}_3\text{O}^+] \)[/tex]:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} \][/tex]
Substitute the values into the formula:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{8.694826047713664 \times 10^{-6}} \][/tex]
After performing the calculation, we get:
[tex]\[ [\text{H}_3\text{O}^+] \approx 1.1501092655705902 \times 10^{-9} \, \text{M} \][/tex]
### Summary
- The concentration of [tex]\( [\text{OH}^-] \approx 8.694826047713664 \times 10^{-6} \, \text{M} \)[/tex]
- The concentration of [tex]\( [\text{H}_3\text{O}^+] \approx 1.1501092655705902 \times 10^{-9} \, \text{M} \)[/tex]
Therefore, the value of [tex]\( [\text{H}_3\text{O}^+] \)[/tex] in the 0.18 M solution of the weak base with [tex]\( K_b = 4.2 \times 10^{-10} \)[/tex] is approximately [tex]\( 1.1501092655705902 \times 10^{-9} \, \text{M} \)[/tex].
