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

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If the $K_{b}$ $K_b$ of a weak base is $1.2 \times 10^{- 6}$ $1.2 times 10^-6$ , 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 $K_{b}$ $K_b$ and its concentration, that is enough. Denote a neutral base as $B$ $B$ . Then the ICE table can be constructed.

$B (a q) + H_{2} O (l) ⇌ {BH}^{+} (a q) + {OH}^{-} (a q)$ $"B"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "BH"^(+)(aq) + "OH"^(-)(aq)$

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

Thus, the $K_{b}$ $K_b$ is:

$K_{b} = \frac{x^{2}}{0.34 - x}$ $K_b = x^2/(0.34 - x)$

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

$K_{b} \approx \frac{x^{2}}{0.34}$ $K_b ~~ x^2/0.34$

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

$x = \sqrt{0.34 K_{b}}$ $x = sqrt(0.34K_b)$

$= 6.39 \times 10^{- 4}$ $= 6.39 xx 10^(-4)$ $M$ $"M"$

Therefore, the $pH$ $"pH"$ is:

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

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

$= 14 - 3.19$ $= 14 - 3.19$

$= 10.81$ $= color(blue)(10.81)$

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

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