# What is the pH of a solution of 750.0 mL of 0.500 M HNO_2 to which 25.00 grams of sodium nitrite has been added (assume no volume change)?

Jun 16, 2017

$\text{pH} = 3.385$

#### Explanation:

The trick here is to recognize the fact that you're dealing with a buffer solution that contains nitrous acid, ${\text{HNO}}_{2}$, a weak acid, and the nitrite anion, ${\text{NO}}_{2}^{-}$, its conjugate base, delivered to the solution by the soluble salt sodium nitrite.

The $\text{pH}$ of a weak acid/conjugate base buffer can be calculated using the Henderson - Hasselbalch equation

"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))

Here

$\text{p} {K}_{a} = - \log \left({K}_{a}\right)$

and ${K}_{a}$ is the acid dissociation constant of the weak acid.

Nitrous acid has an acid dissociation constant equal to

${K}_{a} = 4.0 \cdot {10}^{- 4}$

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

which means that you can write the Henderson - Hasselbalch equation as

"pH" = -log(4.0 * 10^(-4)) + log( (["NO"_2^(-)])/(["HNO"_2]))

"pH" = 3.40 + log( (["NO"_2^(-)])/(["HNO"_2]))

Now, the problem wants you to assume that no volume change occurs when you add the salt, so you can say that

["HNO"_2] = "0.500 M"

In order to find the concentration of the nitrite anion, use the molar mass of sodium nitrite to calculate the number of moles of salt dissolved in the solution.

25.00 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_2/(68.9953color(red)(cancel(color(black)("g")))) = "0.36234 moles NaNO"_2

Sine sodium nitrite dissociates in a $1 : 1$ mole ratio to produce nitrite anions

${\text{NaNO"_ (2(aq)) -> "Na"_ ((aq))^(+) + "NO}}_{2 \left(a q\right)}^{-}$

you can say that the solution will contain $0.36234$ moles of nitrite anions in $\text{750.0 mL}$ of solution.

This implies that the molarity of the nitrite anions, which is calculated by dividing the number of moles of solute by the total volume of the solution in liters, will be

["NO"_2^(-)] = "0.36234 moles"/(750.0 * 10^(-3)color(white)(.)"L") = "0.36234 moles"/0.750 * 1/"1 L"

$= \text{0.48312 moles"/"1 L" = "0.48312 M}$

You can now plug this into the Henderson - Hasselbalch equation to find the $\text{pH}$ of the solution

"pH" = 3.40 + log( (0.48312 color(red)(cancel(color(black)("M"))))/(0.500color(red)(cancel(color(black)("M")))))

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 3.385}}}$

The answer is rounded to three decimal places, the number of sig figs you have for the molarity of the nitrous acid.

Finally, does the result make sense?

Notice that you have

$\left[{\text{NO"_2^(-)] > ["NO}}_{2}\right]$

This tells you that the solution contains more weak acid than conjugate base, which implies that the $\text{pH}$ of the solution will be lower than the $\text{p} {K}_{a}$ of the weak acid.

$\text{pH" = 3.385 < "p"K_a = 3.40" } \to$ makes sense.