# 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?

##### 1 Answer

#### Answer:

#### 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, **weak acid**, and sodium nitrite, **conjugate base**, the nitrite anion,

Now, you're starting with **neutralize** some of the nitrous acid by adding sodium hydroxide, **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

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 **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 **mole** of hydroxide anions **for every mole** of nitrous acid present in solution, and produces **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

#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

#"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**.