# Question 9988f

Dec 26, 2017

Here's what I got.

#### Explanation:

The idea here is that the nitrite anion, ${\text{NO}}_{2}^{-}$, acts as a weak base in aqueous solution, so right from the start, you should expect to find

$\text{pH" > 7" " and " " "pOH} < 7$

The nitrite anions are delivered to the solution by the soluble sodium nitrite in a $1 : 1$ mole ratio, so you know that you start with

["NO"_ 2^(-)]_0 = "0.125 M"

Now, when the nitrite anion is present in aqueous solution, the following equilibrium is established

${\text{NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HNO"_ (2(aq)) + "OH}}_{\left(a q\right)}^{-}$

The base dissociation constant, ${K}_{b}$, of the nitrite anion can be calculated by using the fact that an aqueous solution at ${25}^{\circ} \text{C}$ has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{a} \cdot {K}_{b} = 1 \cdot {10}^{- 14}}}}$

In your case, the acid dissociation of the nitrous acid, ${\text{HNO}}_{2}$, is equal to $6.5 \cdot {10}^{- 4}$, so you can say that the nitrite anion will have

${K}_{b} = \frac{1 \cdot {10}^{- 14}}{6.5 \cdot {10}^{- 4}} = 1.54 \cdot {10}^{- 11}$

Now, by definition, the base dissociation constant is equal to

${K}_{b} = \left(\left[{\text{HNO"_2] * ["OH"^(-)])/(["NO}}_{2}^{-}\right]\right)$

Notice that the nitrite anions react in $1 : 1$ mole ratios to produce nitrous acid and hydroxide anions, so if you say that $x$ $\text{M}$ represents the equilibrium concentrations of the two products, you can say that the equilibrium concentration of the nitrite anions will be equal to

${\left[{\text{NO"_ 2^(-)] = ["NO}}_{2}^{-}\right]}_{0} - x$

This basically means that in order for the reaction to produce $x$ $\text{M}$ of nitrous acid and hydroxide anions, it must consume $x$ $\text{M}$ of nitrite anions.

This means that you have

${K}_{b} = \frac{x \cdot x}{0.125 - x}$

Since the value of the base dissociation constant is so small compared with the initial concentration of the nitrite anions, which means that the equilibrium lies far to the left, you can use the approximation

$0.125 - x \approx 0.125$

You will thus have

${K}_{b} = {x}^{2} / 0.125$

which will get you

x = sqrt( (0.125 * 1.54 * 10^(-11))

$x = 1.39 \cdot {10}^{- 6}$

Since $x$ represents the equilibrium concentration of the hydroxide anions, you can say that you have

["OH"^(-)] = color(darkgreen)(ul(color(black)(1.39 * 10^(-6)color(white)(.)"M")))

Using the fact that an aqueous solution at ${25}^{\circ} \text{C}$ has

color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1 * 10^(-14)color(white)(.)"M"^2)))

you can say that this solution will have

["H"_3"O"^(+)] = (1 * 10^(-14) "M"^color(red)(cancel(color(black)(2))))/(1.39 * 10^(-6)color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(7.19 * 10^(-9)color(white)(.)"M")))

Consequently, you will have

"pOH" = - log(["OH"^(-)])

$\text{pOH} = - \log \left(1.39 \cdot {10}^{- 6}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{5.86}}}$

and

"pH" = - log (["H"_3"O"^(+)])

$\text{pH} = - \log \left(7.19 \cdot {10}^{- 9}\right)$

$\text{pH} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{8.14}}}$

Notice that you get the same result by using

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH + pOH} = 14}}}$

The equilibrium concentrations are rounded to three sig figs, but I'll leave the $\text{pH}$ and the $\text{pOH}$ of the solution rounded to two decimal places instead of three.

As predicted, the $\text{pH}$ of the solution is indeed $> 7$, which is consistent with the fact that the nitrite anion acts as a weak base in aqueous solution. You can also say that sodium nitrite is a basic salt.

$\textcolor{w h i t e}{\frac{a}{a}}$

A MORE ACCURATE APPROACH

An interesting thing to keep in mind here is that pure water already contains hydronium and hydroxide anions $\to$ think the auto-ionization of water here.

At ${25}^{\circ} \text{C}$, pure water has

color(blue)(ul(color(black)(["H"_3"O"^(+)] = ["OH"^(-)] = 1 * 10^(-7)color(white)(.)"M")))

Now, notice that the equilibrium concentration of hydroxide anions that we calculated

["OH"^(-)] = 1.39 * 10^(-6)color(white)(.)"M"

is actually quite small and comparable to the initial concentration of hydroxide anions, i.e. $1 \cdot {10}^{- 7}$ $\text{M}$.

To account for this, you can use the fact that the solution already contains hydroxide anions when you dissolve the salt. So if you use $x$ $\text{M}$ to represent the equilibrium concentration of nitrous acid, you can say that the equilibrium concentration of hydroxide anions will be

$1 \cdot {10}^{- 7} + x$

This means that the concentration of hydroxide anions starts at $1 \cdot {10}^{- 7}$ in pure water and increases by $x$ $\text{M}$ when $x$ $\text{M}$ of nitrite anions react.

The base dissociation constant will now be

${K}_{b} = \frac{x \cdot \left(1 \cdot {10}^{- 7} + x\right)}{0.125 - x}$

Using the same approximation as before, you will end up with

${x}^{2} + 1 \cdot {10}^{- 7} \cdot x - 0.125 \cdot 1.54 \cdot {10}^{- 11} = 0$

This quadratic equation will produce two solutions, one positive and one negative. Since we're looking for concentration here, you can discard the negative solution to say that

$x = 1.34 \cdot {10}^{- 6} \textcolor{w h i t e}{.} \text{M}$

This time, you will have

["OH"^(-)] = 1.34 * 10^(-6)color(white)(.)"M"

["H"_3"O"^(+)] = 7.46 * 10^(-9)color(white)(.)"M"

$\text{pOH} = 5.87$

$\text{pH} = 8.13$

Now, because the percent error that you get by disregarding the initial concentration of hydroxide anions is < color(red)(5%)

"% error" = (1.39 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M"))) - 1.34 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M"))))/(1.34 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M")))) * 100% = 3.73% < color(red)(5%)#

you can use the values that you got without accounting for the initial concentration of the hydroxide anions.