# The #"pH"# at one-half the equivalence point in an acid-base titration was found to be #5.67#. What is the value of #K_a# for this unknown acid?

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The idea here is that at the **half equivalence point**, the

Assuming that you're titrating a weak monoprotic acid **equivalence point**, the strong base will completely neutralize the weak acid.

#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

So when you're adding **equal numbers of moles** of weak acid and of strong base, all the moles of the weak will be consumed and you'll be left with **conjugate base** of the weak acid.

Now, at the **half equivalence point**, you're adding enough moles of the strong base to neutralize **half of the moles** of the weak acid present in the solution.

The reaction will consume **half** of the moles of the weak acid and produce just as many moles of the conjugate base **mole ratios**, meaning that what you consume from the weak acid and the strong base, you produce as the conjugate base.

And so at the **half equivalence point**, the solution will contain **equal numbers of moles** of the weak acid and of its conjugate base, which implies that you're now dealing with a **buffer solution**.

As you know, the **Henderson - Hasselbalch equation**

#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#

At the half equivalence point, you have

#["HA"] = ["A"^(-)]#

which implies that

#log( (["HA"])/(["A"^(-)])) = log(1) = 0#

Therefore, you can say that at the half-equivalence point, the **equal** to the

#color(blue)(ul(color(black)("At the half equivalence point: " -> " pH" = "p"K_a)))#

The **acid dissociation constant** of the weak acid,

#"p"K_a = - log(K_a)#

which implies that you have

#K_a = 10^(-"p"K_a)#

At the **half equivalence point**, you will have

#K_a = 10^(-"pH")#

Plug in your value to find

#K_a = 10^(-5.67) = color(darkgreen)(ul(color(black)(2.1 * 10^(-6))))#

The answer is rounded to two **sig figs**, the number of decimal places you have for the

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