Question

If the hydroxide ion concentration of a 0.109 M NH4OH solution is 0.0014 M, what is the ionization constant for ammonium hydroxide?Group of answer choicesKi = 1.8 × 10-6Ki = 1.8 × 10-5Ki = 1.8 × 10-4Ki = 1.4 × 10-3Ki = 1.3 × 10-2

Answer 1

OPTION B

Explanation

Given information

`\begin{gathered} \lbrack OH^-\rbrack\text{ = 0.109 M} \\ \text{The concentration of a}mmonium\text{ hydrox}ide\text{ solution = 0.0014M} \end{gathered}`

Required? The ionization constant for ammonium hydroxide

The next step is to find the POH

`\begin{gathered} \text{POH = -log}_{10\text{ }}0.0014 \\ \text{POH = 2.85} \end{gathered}`

Recall that,

`\text{POH = }(1)/(2)(\text{pKbase - log C)}`
`\begin{gathered} 2.85\text{ = }(1)/(2)(\text{pKb - log 0.109)} \\ 2.85\text{ }*2\text{ = pKb - log 0.109} \\ 5.7\text{ = pKb - log 0.109} \\ \text{pKb = 5.7 + log 0.109} \\ \text{pkb = 5.7 - 0.96} \\ pkb\text{ = 4.74} \\ \text{pkb = -log kb} \\ kb=10^(-pkb) \\ kb=10^(-4.74) \\ kb\text{ = 1.8 }*10^(-5) \end{gathered}`

Therefore, the correct option is B