# How many milliters of 1.00 M NaOH would you have to add to 200ml of 0.150M HNO_2 to make a buffer with a pH of 4.00?

Aug 1, 2016

$\text{24.1 mL}$

#### Explanation:

Your tool of choice here will be the Henderson-Hasselbalch equation, which for a buffer that consists of a weak acid and its conjugate base looks like this

color(blue)(|bar(ul(color(white)(a/a)"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))

In your case, nitrous acid, ${\text{HNO}}_{2}$, is the weak acid, and sodium nitrite, ${\text{NaNO}}_{3}$ will be the salt of its conjugate base, the nitrite anion, ${\text{NO}}_{2}^{-}$.

Now, you're starting with $\text{200 mL}$ of $\text{0.150 M}$ nitrous acid solution. In order to add some nitrite anions to the solution, you must neutralize some of the nitrous acid by adding sodium hydroxide, $\text{NaOH}$, a strong base.

The net ionic equation that describes this reaction looks like this -- keep in mind that the sodium cations are spectator ions

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

Your goal now will be to use the H-H equation to determine the concentration of conjugate base needed to make this buffer. Nitrous acid has a $\text{p} {K}_{a}$ of $3.39$

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

You will thus have

$4.00 = 3.39 + \log \left(\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right)\right)$

Rearrange to get

$\log \left(\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right)\right) = 4.00 - 3.39$

$\log \left(\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right)\right) = 0.61$

This will be equivalent to

${10}^{\log} \left(\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right)\right) = {10}^{0.61}$

Finally, the ratio that exists between the concentration of conjugate base and the concentration of weak acid is

$\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right) = 4.074$

You will thus have

$\left[\text{NO"_2^(-)] = 4.074 * ["HNO"_2]" "color(orange)("(*)}\right)$

Now, let's assume that the volume of sodium hydroxide you must add to the initial solution is equal to $x$ $\text{mL}$. Use the molarity of the solution to find how many moles of hydroxide anions will be delivered by this volume

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

n_("OH"^(-)) = "1.00 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(x * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

n_("OH"^(-)) = (x/1000)color(white)(a)"moles OH"^(-)

Use the molarity and volume of the nitrous acid solution to calculate how many moles of weak acid it contained

n_("HNO"_2) = "0.150 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(200 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

n_("HNO"_2) = "0.0300 moles HNO"_2

Notice that the neutralization reaction consumes $1$ mole of hydroxide anions for every mole of nitrous acid present in solution, and produces $1$ mole of nitrite anions.

You can thus say that after the reaction takes place, your solution will contain

n_("OH"^(-)) = "0 moles OH"^(-) -> completely consumed

n_("HNO"_2) = "0.0300 moles" - (x/1000)color(white)(a)"moles"

$= \left(\frac{30 - x}{1000}\right) \textcolor{w h i t e}{a} {\text{moles HNO}}_{2}$

n_("NO"_2^(-)) = "0 moles" + (x/1000)color(white)(a)"moles"

$= \left(\frac{x}{1000}\right) \textcolor{w h i t e}{a} {\text{moles NO}}_{2}^{-}$

The total volume of the resulting solution will be

${V}_{\text{total" = "200 mL" + xcolor(white)(a)"mL" = (200 + x)color(white)(a)"mL}}$

The concentrations of the two chemical species in the final solution will be

["HNO"_2] = ( (30-x)/color(red)(cancel(color(black)(1000)))"moles")/((200+x) * color(red)(cancel(color(black)(10^(-3))))"L") = (30-x)/(200+x)color(white)(a)"M"

["NO"_2^(-)] = (x/color(red)(cancel(color(black)(1000)))"moles")/((200+x) * color(red)(cancel(color(black)(10^(-3))))"L") = x/(200+x)color(white)(a)"M"

You can now use equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to find the value of $x$

$\frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{200 + x}}}} = 4.074 \cdot \frac{30 - x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{200 + x}}}}$

This is equivalent to

$x = 4.074 \cdot \left(30 - x\right)$

$x + 4.074 x = 122.22 \implies x = \frac{122.22}{5.074} = 24.09$

Since $x$ represents the volume of the sodium hydroxide solution that must be added to your initial solution of nitrous acid, the answer will be

"volume of NaOH" = color(green)(|bar(ul(color(white)(a/a)color(black)("24.1 mL")color(white)(a/a)|)))

The answer is rounded to three sig figs.