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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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?

Answers

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

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