A solution of ammonia (NH_3) has a pH of 11.22. What is the original molarity of NH_3 in the solution? (Assume K_b for NH_3 is 1.8 × 10^{-5})

• A. 0.10 M
• B. 0.20 M
• C. 1.00 M
• D. 0.50 M

Answer: (B)

Verified step by step guidance

1. Calculate the concentration of hydroxide ions, \([\mathrm{OH}^-]\), from the given pH. First, find the pOH using the relation \(\mathrm{pOH} = 14 - \mathrm{pH}\).
2. Convert the pOH to hydroxide ion concentration using \([\mathrm{OH}^-] = 10^{-\mathrm{pOH}}\).
3. Set up the expression for the base dissociation of ammonia: \(\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}\). Let the initial concentration of ammonia be \(C\), and the change in concentration due to dissociation be \(x = [\mathrm{OH}^-]\).
4. Write the expression for the base dissociation constant \(K_b\) in terms of \(x\) and \(C\): \(K_b = \frac{x^2}{C - x}\). Since \(x\) is small compared to \(C\), approximate \(C - x \approx C\) to simplify the expression to \(K_b \approx \frac{x^2}{C}\).
5. Rearrange the equation to solve for the original molarity \(C\) of ammonia: \(C \approx \frac{x^2}{K_b}\). Substitute the known values of \(x\) and \(K_b\) to find \(C\).