If the $K_b$ of a weak base is $1.2 times 10^-6$ , what is the pH of a 0.34 M solution of this base?

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If the $K_b$ of a weak base is $1.2 times 10^-6$ , what is the pH of a 0.34 M solution of this base?

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Answer 1 · 2017-05-04T21:14:04

It does not matter what base you are looking at. If you know its $K_b$ and its concentration, that is enough. Denote a neutral base as $B$ . Then the ICE table can be constructed.

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

$"I"" ""0.34 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M"$
$"C"" "-x" "" "" "" "-" "" "+x" "" "" "+x$
$"E"" "(0.34 - x)"M"" "-" "" "" "x" "" "" "" "x$

Thus, the $K_b$ is:

$K_b = x^2/(0.34 - x)$

In fact, the $K_b$ is small enough to use the small $x$ approximation. A good rule of thumb is to check whether $K$ is on the order of $10^(-5)$ . Hence, we have:

$K_b ~~ x^2/0.34$

and the equilibrium concentration of $"OH"^(-)$ is readily obtained:

$x = sqrt(0.34K_b)$

$= 6.39 xx 10^(-4)$ $"M"$

Therefore, the $"pH"$ is:

$color(blue)("pH") = 14 - "pOH"$

$= 14 - (-log["OH"^(-)])$

$= 14 - 3.19$

$= color(blue)(10.81)$

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If the $K_b$ of a weak base is $1.2 times 10^-6$ , what is the pH of a 0.34 M solution of this base?

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