Question #a4ab2

Question

Question #a4ab2

1 Answer

Impact of this question

3851 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-28T16:39:17

$p H = 12.22$ $pH=12.22$

Explanation:

In aqueous solution, $N H_{3}$ $NH_3$ reacts with water according to the following reaction:

$N H_{3} (a q) + H_{2} O (l) \to N H_{4}^{+} (a q) + O H^{-} (a q)$ $" " " " " " " "NH_3(aq)+H_2O(l)->NH_4^+(aq)+OH^(-)(aq)$
$I n i t i a l 15 M 0 M 0 M$ $Initial" " " "15M" " " " " " " " " " " "0M" " " " " "0M$
$Change - x + x + x$ $"Change" " " " "-x" " " " " " " " " " " "+x" " " " " "+x$
$E q u i l i b r i u m (15 - x) M + x + x$ $Equilibrium" "(15-x)M " " " " " " ""+x" " " " " "+x$

The equilibrium constant is written as:
$K_{b} = \frac{[N H_{4}^{+}] [O H^{-}]}{N H_{3}} = 1.8 \times 10^{- 5}$ $K_b=([NH_4^+][OH^(-)])/[NH_3]=1.8xx10^(-5)$

Replacing the equilibrium concentrations by their values in the expression of $K_{b}$ $K_b$ :

$K_{b} = \frac{(x) (x)}{(15 - x)} = 1.8 \times 10^{- 5}$ $K_b=((x)(x))/((15-x))=1.8xx10^(-5)$

since the value of $K_{b}$ $K_b$ value is small, we consider $x << 15$ $x"<<"15$

Solving for $x$ $x$ , $x = 1.64 \times 10^{- 2} M$ $x=1.64xx10^(-2)M$

$x$ $x$ represents the concentration of $O H^{-}$ $OH^-$ .
Using the expression of $K_{w} = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$ $K_w=[H^+][OH^-] =1.0xx10^(-14)$
$[H^{+}] = \frac{K_{w}}{[O H^{-}]} = \frac{1.0 \times 10^{- 14}}{1.64 \times 10^{- 2}} = 6.09 \times 10^{- 13} M$ $[H^+]=(K_w)/([OH^-])=(1.0xx10^(-14))/(1.64xx10^(-2))=6.09xx10^(-13)M$

Then, the pH is calculated by:
$p H = - log ([H^{+}])$ $pH=-log([H^+])$ $\Rightarrow p H = - log (6.09 \times 10^{- 13})$ $=> pH=-log(6.09xx10^(-13))$

$p H = 12.22$ $color(blue)(pH=12.22)$

Science

Math

Humanities

... and beyond

Question #a4ab2

1 Answer

Explanation:

Related questions

Impact of this question