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 :

Answer : The value of [tex]pK_a[/tex] of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

[tex]\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol[/tex]

Now we have to calculate the value of [tex]pK_a[/tex] of the weak acid.

The equilibrium chemical reaction is:

                          [tex]HA+KOH\rightleftharpoons HK+H_2O[/tex]

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[HK]}{[HA]}[/tex]

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

[tex]3.85=pK_a+\log (\frac{0.03}{0.22})[/tex]

[tex]pK_a=4.72[/tex]

Therefore, the value of [tex]pK_a[/tex] of the weak acid is, 4.72

whitneytr12

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:

[tex] pH = pKa + log(\frac{[KA]}{[HA]}) [/tex]  (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.

[tex] n_{KOH} = C_{KOH}*V_{KOH} = 3.00 \:mol/L*0.010 \:L = 0.03 \:moles [/tex]

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

[tex] n_{a} = n_{i} - n_{KOH} = 0.25 \:moles - 0.03 \: moles = 0.22 \:moles [/tex]

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

[tex] [HA] = \frac{n_{a}}{V} = \frac{0.22\:moles}{1.5 L} = 0.15\: mol/L [/tex]

[tex] [KA] = \frac{0.03 \:moles}{1.5 L} = 0.02 \:mol/L [/tex]

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

[tex] 3.85 = pKa + log(\frac{0.02}{0.15}) [/tex]

[tex] pKa = 4.72 [/tex]

Therefore, the pKa of the weak acid is 4.72.            

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