Respuesta :
Answer:
[tex]\boxed{2.64}[/tex]
Explanation:
The equation for the equilibrium is
NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
1. Set up an ICE table
[tex]\begin{array}{ccccccc}\text{NH$_{4}^{+}$} & + & \text{H$_{2}$O}& \, \rightleftharpoons \, & \text{NH$_{3}$} & + & \text{H$_{3}$O$^{+}$}\\0.289 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.289-x & & & &x & & x \\\end{array}[/tex]
2. Solve for x
[tex]K_{\text{c}} = \dfrac{\text{[NH$_{4}^{+}$][OH$^{-}$]}}{\text{[NH$_3$]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.289-x} = 1.8 \times 10^{-5}\\\\\text{Check for negligibility of }x\\\\\dfrac{0.289 }{1.8 \times 10^{-5}} = 16 000 > 400\\\\\therefore x \ll 0.289\\\\\begin{array}{rcl}\\\dfrac{x^{2}}{0.289}& = & 1.8 \times 10^{-5}\\\\x^{2}& = & 0.289 \times1.8 \times 10^{-5}\\x^{2}& = & 5.202 \times 10^{-6}\\x & = &2.281 \times 10^{-3}\\\end{array}\\[/tex]
3. Calculate the pH
[tex]\text{[H$_{3}$O$^{+}$]}= x \text{ mol$\cdot$L$^{-1}$} = 2.281 \times 10^{-3} \text{ mol$\cdot$L$^{-1}$}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{2.281 \times 10^{-3}} = \boxed{\mathbf{2.64}}\\\text{The pH of the solution is } \boxed{\textbf{2.64}}[/tex]
The pH of a 0.289 M solution of ammonium chloride at 25 degree C is equal to 2.64.
How do we calculate the pH?
pH of any solution will be calculated as:
pH = -log[H⁺]
Given chemical reaction in equilibrium with ICE table will be represented as:
NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
Initial: 0.289 0 0
Change: -x +x +x
Equilibrium: 0.289-x x x
Equation for Kb for this reaction is:
Kb = [NH₃][H₃O⁺] / [NH₄⁺]
Given value of Kb = 1.8 × 10⁻⁵
On putting values equation becomes
1.8 × 10⁻⁵ = x² / 0.289-x
Value of x is negligible as compared to 0.289 so equation becomes
1.8 × 10⁻⁵ = x² / 0.289
x = 2.28 × 10⁻³
i.e. [NH₃] = [H₃O⁺] = 2.28 × 10⁻³
Now we calculate the pH by putting this concentration value on the equation & we get,
pH = -log(2.28 × 10⁻³)
pH = 2.64
So, resultant pH is 2.64.
To know more about pH, visit the below link:
https://brainly.com/question/8758541
