Titration problem?
A weak acid HA is titrated with a standard base ( NaOH solution) and the following titration curve is obtained.
For a weak acid , the dissociation constant , Ka , is defined as:
HA + H2O <=> #A^- + H3O^+#
Ka= [A-][H3O+]/[HA]
The realtionship between Ka and pH is :
-lgKa=pH - lg{[A-]/[HA]}
and pH = -lg[H3O+]
Using the information provided determine the dissociation constant ,Ka , for the weak acid HA.
a) #10^{-4,2}#
b)#10^{-4,5}#
c)#10^{-7,8}#
d)#10^{-11,5}#
A weak acid HA is titrated with a standard base ( NaOH solution) and the following titration curve is obtained.
For a weak acid , the dissociation constant , Ka , is defined as:
HA + H2O <=>
Ka= [A-][H3O+]/[HA]
The realtionship between Ka and pH is :
-lgKa=pH - lg{[A-]/[HA]}
and pH = -lg[H3O+]
Using the information provided determine the dissociation constant ,Ka , for the weak acid HA.
a)
b)
c)
d)
2 Answers
Explanation:
The
Sodium hydroxide
One formula unit of sodium hydroxide reacts with one
Appearantly
Substituting
Therefore halfway to that point at the titration midpoint at which
Additionally, given that
given that
Therefore
Reference
"Titration Fundamentals", Chemistry LibreText, https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_Fundamentals
See also
"Titration of a Weak Base with a Strong Acid", https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Base_with_a_Strong_Acid
Explanation:
As an alternative approach, you can use the fact that at the half-equivalence point, exactly half of the weak acid has been neutralized by the strong base.
The chemical equation that describes this neutralization reaction looks like this
#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#
At the half-equivalence point, you add enough moles of strong base, which I've represented here by
Since the reaction consumes the weak acid and produces the conjugate base
#["HA"] = ["A"^(-)]#
Now, notice that at the equivalence point,
So if it takes
This means that the half-equivalence point corresponds to
The question provides you with the equation
#-"lg"(K_a) = "pH" - "lg"( (["A"^(-)])/(["HA"]))#
which is a version of the Henderson - Hasselbalch equation that can be used to calculate the
Plug in the
#- "lg"(K_a)= 4.5 - "lg"(1)#
This gets you
#"lg"(K_a) = - 4.5#
which implies
#color(darkgreen)(ul(color(black)(K_a = 10^(-4.5))))#
See this answer for a similar question.
SIDE NOTE: The notation used here is based on the fact that
#"lg"(x) = log_(10)(x)#
See here for more info on that.