Question

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?

Answer 1

`"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, `"HNO"_2`, is the weak acid, and sodium nitrite, `"NaNO"_3` will be the salt of its conjugate base, the nitrite anion, `"NO"_2^(-)`.

Now, you're starting with `"200 mL"` of `"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, `"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

`"HNO"_ (2(aq)) + "OH"_ ((aq))^(-) -> "NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l))`

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 `"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( (["NO"_2^(-)])/(["HNO"_2]))`

Rearrange to get

`log( (["NO"_2^(-)])/(["HNO"_2])) = 4.00 - 3.39`

`log( (["NO"_2^(-)])/(["HNO"_2])) = 0.61`

This will be equivalent to

`10^log( (["NO"_2^(-)])/(["HNO"_2])) = 10^0.61`

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

`(["NO"_2^(-)])/(["HNO"_2]) = 4.074`

You will thus have

`["NO"_2^(-)] = 4.074 * ["HNO"_2]" "color(orange)("(*)")`

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

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(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"`

`= ((30-x)/1000)color(white)(a)"moles HNO"_2`

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

`=(x/1000)color(white)(a)"moles NO"_2^(-)`

The total volume of the resulting solution will be

`V_"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 `color(orange)("(*)")` to find the value of `x`

`x/color(red)(cancel(color(black)(200 + x))) = 4.074 * (30-x)/color(red)(cancel(color(black)(200 + x)))`

This is equivalent to

`x = 4.074 * (30-x)`

`x + 4.074x = 122.22 implies x= 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.