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45 votes
45 votes
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

User Gambit Support
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1 Answer

25 votes
25 votes

ANSWER

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

User Ivan Skoric
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