What happens to the hydroxide ion concentration when you add caustic soda to ammonia?

Question

Question.

Find the $[\ce{OH-}]$ and the pH of a $\pu{0.20 M}$ $\ce{NH3}$. If given $\pu{1 L}$ of the previous solution, when adding $\pu{100 mL}$ of $\pu{0.2 M}$ $\ce{NaOH}$, what would be the new pH? Given $K_\mathrm b=1.8 \times 10^{-5}$

Attempt. From the info of the constant

$$K_\mathrm b=\frac{x^2}{0.2-x}\implies x=\pu{3.79 \times 10^-3 M}$$

Hence from the expression of the pH:

$$\mathrm{pH}=14+\log(x)=11.58$$

But for the second part I'm having issues to differ that when something is added, what I have to do is:

• Since after the addition of a base, you are adding more concentration of $\ce{OH-}$, hence by Le Chateliers' principle, the system will move to the left, considering the formula is

$$\ce{NH3 +H2O -> NH4+ +OH-}$$

hence, the $[\ce{OH-}] = x-y$

• Since it's just an addition, do $[\ce{OH-}]= x+y$, ignoring Le Chateliers' Principle

Pretty lost when it comes to when to use which bullet point, of course in both you'd use that $K_\mathrm b$ is constant after the addition of anything but still unsure.

Answer 1

The initial system contains $0.0038$ mol $\ce{OH-}$ (and $\ce{NH4^+}$) and $0.1962$ mol $\ce{NH3}$. Adding $0.02$ mol $\ce{NaOH}$ will consume $y$ mol of $\ce{NH4^+}$ and of $\ce{OH-}$ and produce the same $y$ moles of $\ce{NH3}$. As a result, the final amounts are :

$\ce{n(OH^-) = 0.02 mol + 0.0038 mol - y = 0.0238 mol - y}$ ;

$\ce{n(NH3) = 0.1962 mol + y}$ ;

$\ce{n(NH4^+) = 0.0038 mol - y}$.

Forgetting the units, the final expression for the equilibrium constant is $$\ce{K_b = \frac{[NH4^+][OH-]}{[NH3]}} = \frac{\frac{0.0038 - y}{1.1} ·\frac{0.0238 - y}{1.1}}{\frac{0.1962 + y}{1.1}} = \frac{(0.0038 - y)(0.0238 - y)}{(0.1962 + y)·1.1 }= {1.8·10^{-5}}$$ This is a $2$nd degree equation, and its reasonable solution is : $\ce{y = 0.0036}$. So $\ce{[OH^-] = 0.0202 M}$.

As a consequence, $\ce{[H^+] = 4.95 10^{-13}}$. And $\ce{pH = 12.305}$.

If we had ignored the presence of $\ce{NH3}$ in the initial solution, and added $100$ mL of a $0.02$ M $\ce{NaOH}$ in $1$ L pure water, we would have obtained nearly the same pH = $12.301$ ! Surprisingly enough, the presence of a large excess of $\ce{NH3}$ in solution has a negligible effect on the pH of the solution.

Answer 2

I have added below the general solution, defining and solving a set of algebraic equations by the substitution method, as hinted in my comment before. But in this case, it is rather a cannon against a sparrow.

Enough is to replace in your equation $x^2$ by $x(x + [\ce{NaOH}\text{, final}]$, where $x$ is additional $[\ce{OH-}]=[\ce{NH4+}]$ due $\ce{NH3}$.

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The general equation set approach:

There are often less standard situations, like this with $\ce{NaOH}$ addition, without handy equations to be used. In such a case, a generally applicable approach using known, generally valid equations can be applied.

Let perform the notation simplification to have easier writing:

$x=[\ce{H+}], y=[\ce{OH-}], z=[\ce{NH3}], r=[\ce{NH4+}], \\s=[\ce{Na+}], a=c(\ce{NH3},\mathrm{total}), b=c(\ce{NaOH}) $

2 equilibrium equations:

$$K_\mathrm{w}=xy$$ $$K_\mathrm{b}=\frac{yr}{z}$$

2 concentration inventories:

$$z + r = a$$ $$s = b$$

1 charge balance:

$$x + r + s = y$$

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substitution for s:

$$K_\mathrm{w}=xy$$ $$K_\mathrm{b}=\frac{yr}{z}$$ $$z + r = a$$ $$x + r + b = y$$

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substitution for x:

$$K_\mathrm{b}=\frac{yr}{z}$$ $$z + r = a$$ $$\frac{K_\mathrm{w}}{y} + r + b = y$$

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substitution for z:

$$r(\frac{y}{K_\mathrm{b}} + 1) = a$$ $$\frac{K_\mathrm{w}}{y} + r + b = y$$

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substitution for r:

$$\frac{K_\mathrm{w}}{y} + \frac{a}{\frac{y}{K_\mathrm{b}} + 1} + b = y$$

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The first term represent [H+] in charge balance equation. Its value is roughly $\pu{E-12}$ which can be neglected: $$ \frac{a}{\frac{y}{K_\mathrm{b}} + 1} + b = y$$

$$ aK_\mathrm{b} + b(y + K_\mathrm{b}) = y(y + K_\mathrm{b})$$

$$ aK_\mathrm{b} + by + bK_\mathrm{b} = y^2 + yK_\mathrm{b}$$

$$ y^2 + y(K_\mathrm{b}-b)- K_\mathrm{b}(a + b) = 0$$

$$y = \frac{-K_\mathrm{b}+b \pm \sqrt{{(K_\mathrm{b}-b)}^2 +4K_\mathrm{b}(a + b)}}{2}$$

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as $K_\mathrm{b} = \pu{1.8e-5}$, $a = 0.2 \frac {10}{11} \mathrm{M}$, $b=0.2 \frac {1}{11} \mathrm{M}$.....

$$y \approx \frac{-\pu{1.8e-5} + 0.2 \frac {1}{11} \pm \sqrt{{(\pu{1.8e-5}-0.2 \frac {1}{11})}^2 +4 \cdot \pu{1.8e-5} \cdot 0.2 }}{2}$$

$$y \approx \frac{0.018164 \pm \sqrt{0.018164^2 + \pu{1.44e-5} }}{2}$$

$$y \approx 0.01836$$, therefore,

$$[\ce{OH-}] \approx \pu{0.01836 mol L-1}$$

$$\mathrm{pH} = 14 + \log{0.01836} \approx 12.26$$

If we compare it to $\ce{OH-}$ of $\ce{NaOH}$ origin:

$$[\ce{OH-}] = \pu{0.2 \frac {1}{11} mol L-1} \approx \pu{0.01818 mol L-1} $$, $$\mathrm{pH} = 14 + \log{0.01818} \approx 12.26$$

(Without corrections for activity coefficients, it does not make sense for higher pH precission.)

so for $\ce{OH-}$ coming from $\ce{NH3}$:

$$[\ce{OH-}]\approx \pu{0.00018 mol L-1}$$