Question #f5fc3

1 Answer
Mar 31, 2016

Answer:

#"pH" = 4.04#

Explanation:

Sodium hydroxide, #"NaOH"#, a strong base, will react with lactic acid, #"C"_3"H"_6"O"_3#, a weak acid, to form water and aqueous sodium lactate, #"NaC"_3"H"_5"O"_3#.

The sodium cations, #"Na"^(+)#, will act as spectator ions in this reaction, so you can exclude them from the balanced chemical equation, which will look like this

#"C"_ 3"H"_ 6"O"_ (3(aq)) + "OH"_ ((aq))^(-) -> "C"_ 3"H"_ 5"O"_ (3(aq))^(-) + "H"_ 2"O"_((l))#

Notice that the two reactants react in a #1:1# mole ratio. This tells you that the reaction will always consume equal number of moles of lactic acid and of hydroxide anions, #"OH"^(-)#, delivered to the reaction by the sodium hydroxide solution.

Your goal now is to use the molarities and volumes of the two solutions to determine how many moles of each you're delivering to the reaction.

#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)|)))#

Do not forget that the volume of the solution must be expressed in liters, so make sure to convert the volumes of the two solutions from milliliters to liters

#"1 L" = 10^3"L"#

You will have

#n_"lactic acid" = "0.10 mol" color(red)(cancel(color(black)("L"^(-1)))) * 25.0 * 10^(-3)color(red)(cancel(color(black)("L")))#

#= " 0.00250 moles lactic acid"#

#n_(OH^(-)) = "0.15 mol" color(red)(cancel(color(black)("L"^(-1)))) * 10.0 * 10^(-3)color(red)(cancel(color(black)("L")))#

# = " 0.00150 moles OH"^(-)#

So, you have more moles of weak acid than of strong base, which means that the hydroxide anions will be completely consumed by the reaction.

Another thing to notice is that you have #1:1# mole ratios between both reactants and the lactate anion, the conjugate base of the lactic acid.

This means that for every mole of each reactant consumed by the reaction, you get one mole of lactate anions.

After the neutralization reaction is complete, your solution will contain

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

#n_"lactic acid" = 0.00250 - 0.00150 = "0.00100 moles C"_3"H"_6"O"_3#

#n_"lactate ions" = 0 + 0.00150 = "0.00150 moles C"_3"H"_5"O"_3^(-)#

The total volume of the solution will be

#V_"total" = "25.0 mL" + "15.0 mL" = "40.0 mL"#

At this point, you should be able to recognize the fact that you've created a buffer solution that contains a weak acid and its conjugate base in comparable amounts.

As you know, the pH of a buffer solution can be calculated using the Henderson - Hasselbalch equation

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

Lactic acid has a #pK_a# of #3.86#

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

The concentration of the weak acid and conjugate base will be

#["C"_3"O"_6"O"_3] = "0.00100 moles"/(40.0 * 10^(-3)"L") = "0.0250 M"#

#["C"_3"H"_5"O"_3^(-)] = "0.00150 moles"/(40.0 * 10^(-3)"L") = "0.0375 M"#

Since the solution contains more conjugate base than weak acid, you can expect its pH to be greater than the #pK_a# of the acid.

Plug in your values to get

#"pH" = 3.86 + log( (0.0375 color(red)(cancel(color(black)("M"))))/(0.0250color(red)(cancel(color(black)("M")))))#

#"pH" = color(green)(|bar(ul(color(white)(a/a)4.04color(white)(a/a)|)))#