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I'am working on this exercice, Is my answer correct? and can you please help me to solve the second part?

exercice:

1/Calculate the ionization coefficient and pH (with two methods) of a 0.1M ammonia solution of pkb = 4.7

2/ to one liter of the previous solution, an equal volume of water is added. Calculate the new ionization coefficient and deduce the pH of this new solution


1/ For the first part my answer is :

$\ce{NH3 + H2O = NH4+ + OH^-}\tag{1}$

$Ka = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3+]}}\tag{2}$

The coefficient of ionization $\alpha$

$\alpha = \sqrt{Ka/C°}\tag{3}$

$Pkb = \log{kb}\tag{4}$

$Kb= 10-pkb\tag{5}$

$\alpha =0,014\tag{6}$

Two methods:

First one:

$pH= \frac{1}{2} (pka + 14 + \log{C°})\tag{7} $

$pka + pkb=14\tag{8}$

$pH = 11,15\tag{9}$

second method:

$14 + \log (\alpha C°) = 11,15\tag{10}$

for the second part, i am lost

Thank you

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  • $\begingroup$ I tried to edit this to make sense of it, but I left the OP's mistakes and sloppy terminology. Also note that OP uses a comma for a decimal point which I left in. $\endgroup$ – MaxW Jul 22 '18 at 15:23
  • $\begingroup$ Trying to clear up a terminology problem. Is this problem from a book written in English or did you translate it? I'm assuming that ionization coefficient was mean to mean the fraction of the ammonia that ionizes. The common term for that in English is ionization degree. $\endgroup$ – MaxW Jul 23 '18 at 22:12
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The general method to solve problems of this type can be found in the answer here, How to set up equation for buffer reaction?. In your case you need to modify this equation. To follow the text below you need to look at the main result in the linked answer (rather than repeat all this here). The derivation is straightforward although the equation is a bit complicated.

The base constant $K_\mathrm{B}$ is given but as $K_\mathrm{w}=K_\mathrm{A}K_\mathrm{B}$ we can start with main eqn in linked page and convert to $k_\mathrm{B}$ later on. In this case as ammonia is a base (with concentration $c_b= 0.1$ M) we may suppose that any acid concentration $c_\mathrm{a} \approx 0$ then,

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} + \mathrm{[H^+]} - \mathrm{[OH^-])} }{(- \mathrm{[H^+]} + \mathrm{[OH^-]} )} $$

Additionally we can reasonably assume that $\mathrm{[OH^-] \gg [H^+]}$ and so

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} - \mathrm{[OH^-])} }{ \mathrm{[OH^-]} } $$

This can be solved by substituting for $K_\mathrm{A}=K_\mathrm{w}/K_\mathrm{B}$ and for $\ce{[H^+]}$ as $K_\mathrm{w}=\ce{[H^+][OH^-]}$ and solve the quadratic equation. But as $c_\mathrm{b} \gg \ce{[OH^-]}$ we can simplify by letting $c_\mathrm{b} - \ce{[OH^-]} \to c_\mathrm{b}$ then the equation becomes

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{c_\mathrm{b} }{ \mathrm{[OH^-]} } $$

We can now get

$$\frac{K_\mathrm{w}}{K_\mathrm{B}}= \frac{K_\mathrm{w}c_\mathrm{b}} { \mathrm{[OH^-]^2}} $$

from which

$$\ce{[OH]^- } =\sqrt{K_\mathrm{b}c_\mathrm{b}}$$

and so you can obtain the pH.

When you dilute, the concentration of base $c_\mathrm{b}$ is halved.

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The first part of the question ask for the calculation of the pH two different ways and the ionization coefficient for a 0.1 molar ammonia solution. Also given that $pK_\beta = 4.7$.

First calculation of pH using $K_\beta$ of ammonia

$K_\beta = 10^{-pK_\beta} = 2.00\times10^{-5}\tag{1}$

Note: I'm using two extra significant figures for intermediate calculations. The answer should only have one since only one is given $pK_\beta$.

$\ce{NH3 + H2O <=> NH4+ + OH^-}\tag{2}$

$K_\beta = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3]}}\tag{3}$

Little of the ammonia will ionize to form the ammonium cation, but enough will that the ionization of water itself doesn't need to be considered. So $\ce{[NH4^+] = [OH-] }$ and $\ce{[NH3] = 0.1}$ molar. So

$K_\beta = \dfrac{\ce{[OH^-]^2}}{\ce{[NH3]}}\tag{4}$

$\ce{[OH-]} =\sqrt{\ce{K_\beta[NH3]}} = \sqrt{(2.0\times10^{-5})(0.1)} = 1.414\times10^{-3}\tag{5}$

!!Check!!

We assumed that little of the ammonia would be ionized. There was 0.1 moles originally and $1.414\times10^{-3}$ did ionize. So there is 0.098586 moles of ammonia left. So the assumption is reasonable, especially since the given $pK_\beta$ only had one significant figure.

$\ce{[H^+]} = \dfrac{K_w}{\ce{[OH^-]}}= \dfrac{1.0\times10^{-14}}{1.414\times10^{-3}} = 7.072\times10^{-12}\tag{6}$

$pH = - \log{\ce{[H^+]}} = - \log{7.072\times10^{-12}} = 11.1504 \ce{->[rounding(1)] = 11.2}\tag{7}$

Second calculation of pH using $K_\beta$ of ammonia

For this calculation let $x$ be the molarity of $\ce{NH4^+}$. Again assume that autoionization of water is negligible. Hence $\ce{[OH^-]} = x$ also, and $\ce{[NH3] = 0.1 - x}$. Substituting into equation (3) yields:

$2.0\times10^{-5} = \dfrac{x^2}{0.1 - x}\tag{8}$

rearranging we get the quadratic equation

$x^2 + (2.0\times10^{-5})x- 2.0\times10^{-6} \tag{9}$

for which the real solution is $x = 1.424\times10^{-3} $. Calculating the pH using method given above yields $pH = 11.2$ again.

Calculation of pH using $K_\alpha$ of ammonium

$pK_\alpha = 14.0 - pK_\beta= 14.0 - 4.7 = 9.3\tag{10}$

$K_\alpha = 10^{-pK_\alpha} = 10^{-9.3} = 5.012\times10^{-10}\tag{11}$

$\ce{NH4^+ <=> NH3 + H^+}\tag{12}$

$K_\alpha = \dfrac{\ce{[NH3][H^+]}}{\ce{[NH4^+]}}\tag{13}$

but $\ce{[H+]} = \dfrac{K_w}{\ce{[OH^-]}}$ so

$K_\alpha = \dfrac{\ce{[NH3]K_w}}{\ce{[NH4^+][OH^-]}}\tag{14}$

Again $\ce{[NH4^+] = [OH-] }$ and $\ce{[NH3] = 0.1}$ molar. So rearranging and solving for $\ce{[OH^-]}$ gives

$\ce{[OH^-]} = \sqrt{\dfrac{\ce{[NH3]K_w}}{K_a}} = \sqrt{\dfrac{(0.1)(1.0\times10^{-14})}{5.012\times10^{-10}}} = 1.413\times10^{-3}\tag{15}$

Solving for pH as above will yield 11.2.


Solving for the "ionization coefficient"

I don't remember encountering the term ionization coefficient ($\alpha$) before so I'm assuming it means the fraction of the ammonia which is ionized. Let $C_N$ be the nominal concentration of ammonia. Then the mass balance of ammonia species yields:

$\ce{C_N = [NH3] + [NH4^+]}\tag{16}$

so the desired calculation is:

$\alpha = \dfrac{[NH4^+]}{\ce{C_N}}\tag{17}$

Solving equation (2) for $\ce{[NH3]}$ and substituting into equation (16) yields

$\ce{C_N} = \dfrac{\ce{[NH4^+][OH^-]}}{K_\beta} + \ce{[NH4^+]}\tag{18}$

$\ce{K_\beta C_N = [NH4^+][OH^-] + [NH4]K_\beta}\tag{19}$

$\dfrac{\ce{[NH4^+]}}{C_N} = \dfrac{K_\beta}{\ce{[OH^-]} + K_\beta} = \dfrac{2.000\times10^{-5}}{1.414\times10^{−3} + 2.000\times10^{-5}} = 1.4\%\tag{20}$


For the second part the 0.1M ammonia solution has been diluted to 0.05 molar. Just redo calculations above.

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  • $\begingroup$ Maxw: According to the charge balance equation:$$\ce{[H+] +[NH4+]= [OH-]}$$ Why we assuming $\ce{[NH4+]= [OH-]}$ $\endgroup$ – Adnan AL-Amleh May 30 at 17:55
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    $\begingroup$ @AdnanAL-Amleh - $\ce{[NH4^+] \gg [H^+]}$ $\endgroup$ – MaxW May 30 at 18:00

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