When a general base associates in water, with $K_b$ $"<<"$ $10^(-5)$ , what is the algebraic expression for $x$ as a function of $K_b$ and the equilibrium concentration of the base?

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When a general base associates in water, with $K_b$ $"<<"$ $10^(-5)$ , what is the algebraic expression for $x$ as a function of $K_b$ and the equilibrium concentration of the base?

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Answer 1 · 2017-03-09T08:05:31

As usual, write out the reaction.

$"B"(aq) + "H"_2"O"(l) rightleftharpoons "OH"^(-)(aq) + "BH"^(+)(aq)$

By constructing an ICE table, you would obtain:

$K_b = (["OH"^(-)]["BH"^(+)])/(["B"]) = x^2/(["B"] - x)$

As $K_b$ $"<<"$ $10^(-5)$ , we can of course use the small $x$ approximation, and thus, $["B"] - x ~~ ["B"]$ . Thus:

$color(blue)(x ~~ sqrt(K_b["B"]))$

And you can always use this expression for a weak base equilibrium in which $K_b$ is on the order of $10^(-5)$ or less. What does $x$ represent, and what is the answer?

Did you check your answer by trying to see if the $"pH"$ was more than $7$ ?

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When a general base associates in water, with $K_b$ $"<<"$ $10^(-5)$ , what is the algebraic expression for $x$ as a function of $K_b$ and the equilibrium concentration of the base?

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When a general base associates in water, with K_bK_b "<<""<<" 10^(-5)10^(-5), what is the algebraic expression for xx as a function of K_bK_b and the equilibrium concentration of the base?

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When a general base associates in water, with $K_b$ $"<<"$ $10^(-5)$ , what is the algebraic expression for $x$ as a function of $K_b$ and the equilibrium concentration of the base?