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?
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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.
Answer : The value of [tex]pK_a[/tex] of the weak acid is, 4.72Explanation :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 0At eqm. (0.25-0.03) 0.03 0.03 = 0.22Using 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