Question

A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was diluted to 1.500 L. The measured pH of the solution was 3.85. What is the pKa of the weak acid?

Answer 1

Answer : The value of
`pK_a` of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

`\text{Moles of }KOH=\text{Concentration of }KOH* \text{Volume of solution}`

`\text{Moles of }KOH=3.00M* 10.0mL=30mmol=0.03mol`

Now we have to calculate the value of
`pK_a` of the weak acid.

The equilibrium chemical reaction is:

`HA+KOH\rightleftharpoons HK+H_2O`

Initial moles 0.25 0.03 0

At eqm. (0.25-0.03) 0.03 0.03

= 0.22

Using Henderson Hesselbach equation :

`pH=pK_a+\log ([Salt])/([Acid])`

`pH=pK_a+\log ([HK])/([HA])`

Now put all the given values in this expression, we get:

`3.85=pK_a+\log ((0.03)/(0.22))`

`pK_a=4.72`

Therefore, the value of
`pK_a` of the weak acid is, 4.72

Answer 2

The pKa of the 0.25 mol sample of a weak acid calculated after its reaction with 10.0 mL of 3.00 M KOH is 4.72.

When the weak acid reacts with KOH, we have:

HA(aq) + KOH(aq) ⇄ H₂O(l) + KA(aq) (1)

The pKa of the reaction above can be calculated with the Henderson-Hasselbalch equation:

`pH = pKa + log(([KA])/([HA]))` (2)

Where:

pH = 3.85

[HA]: is the concentration of the weak acid

[KA]: is the concentration of the salt

To find the pKa, we need to calculate the values of [HA] and [KA].

First, let's find the number of moles of KOH.

`n_(KOH) = C_(KOH)*V_(KOH) = 3.00 \:mol/L*0.010 \:L = 0.03 \:moles`

Now, when the weak acid reacts with KOH, the number of moles of the acid that remains in the solution is:

`n_(a) = n_(i) - n_(KOH) = 0.25 \:moles - 0.03 \: moles = 0.22 \:moles`

When the resulting solution is then diluted to 1.500 L, the concentration of the HA and KA is:

`[HA] = (n_(a))/(V) = (0.22\:moles)/(1.5 L) = 0.15\: mol/L`

`[KA] = (0.03 \:moles)/(1.5 L) = 0.02 \:mol/L`

After entering the values of pH, [HA], and [KA] into equation (2), we have:

`3.85 = pKa + log((0.02)/(0.15))`

`pKa = 4.72`

Therefore, the pKa of the weak acid is 4.72.