A 1.50 L buffer solution is 0.250 M HF and 0.250 M NaF. What is the pH of the solution after the addition of 0.0500 moles of solid NaOH? Assume no volume change. Ka for HF is 3.5x10^-4
1 Answer
Explanation:
Your buffer contains hydrofluoric acid,
When the sodium hydroxide solution is added, assuming with no change in the total volume of the buffer, you can expect the weak acid and the strong base to neutralize each other.
Moreover, this reaction will result in the production of more conjugate base.
So, the balanced chemical equation for the reaction between hydrofluoric acid and sodium hydroxide looks like this
"HF"_text((aq]) + "NaOH"_text((aq]) -> "NaF"_text((aq]) + "H"_2"O"_text((l])HF(aq]+NaOH(aq]→NaF(aq]+H2O(l]
Notice that
Use the molarity of the hydrofluoric acid and the volume of the buffer to determine how many moles you have in solution
color(blue)(c = n/V implies n = c * V)c=nV⇒n=c⋅V
n_(HF) = "0.250 M" * "1.5 L" = "0.375 moles HF"nHF=0.250 M⋅1.5 L=0.375 moles HF
Do the same for the conjugate base
n_(F^(-)) = "0.250 M" * "1.5 L" = "0.375 moles F"^(-)nF−=0.250 M⋅1.5 L=0.375 moles F−
Now, you're adding
The number of moles of hydrofluoric acid will be changed to
n_(HF) = "0.375 moles" = "0.0500 moles" = "0.325 moles"nHF=0.375 moles=0.0500 moles=0.325 moles
The number of moles of fluoride anions will increase by the same amount
n_(F^(-)) = "0.375 moles" + "0.0500 moles" = "0.425 moles F"^(-)nF−=0.375 moles+0.0500 moles=0.425 moles F−
Use the volume of the buffer to calculate the new molarities of the weak acid and its conjugate base
["HF"] = "0.325 moles"/"1.5 L" = "0.21667 M"[HF]=0.325 moles1.5 L=0.21667 M
["F"^(-)] = "0.425 moles"/"1.5 L" = "0.28333 M"[F−]=0.425 moles1.5 L=0.28333 M
Finally, use the Henderson - Hasselbalch equation to find the pH of the buffer
color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))pH=pKa+log([conjugate base][weak acid])
Use the acid dissociation constant,
pK_a = - log(K_a)pKa=−log(Ka)
The pH of the solution will thus be
"pH" = -log(K_a) + log( (["F"^(-)])/(["HF"]))pH=−log(Ka)+log([F−][HF])
"pH" = -log(3.5 * 10^(-4)) + log( (0.28333color(red)(cancel(color(black)("M"))))/(0.21667color(red)(cancel(color(black)("M")))))
"pH" = color(green)(3.57)
Finally, does this result make sense?
Notice that your starting buffer had equal concentrations of weak acid and conjugate base. This means that the H-H equation can be reduced to
"pH" = pK_a + log(1) = pK_a
Initially, the pH of the solution was equal to
The buffer then converts the strong base to weak base, and so the concentration of conjugate base is bigger than that of the weak acid. This is why the pH of the buffer increases after the addition of the strong base.
