A solution of weak acid HA was titrated with base NaOH. The equivalent point was reached when 40 ml. of 0.1 M NaOH has been added. Now 20 ml. of 0.1 M HCl were added to titrated solution, the pH was found to be 5.0. The value of pKa of HA is :
Question
A solution of weak acid HA was titrated with base NaOH. The equivalent point was reached when 40 ml. of 0.1 M NaOH has been added. Now 20 ml. of 0.1 M HCl were added to titrated solution, the pH was found to be 5.0. The value of pKa of HA is :
Solution
To solve this problem, we need to understand that at the equivalence point, all of the weak acid (HA) has been neutralized by the base (NaOH) to form its conjugate base (A-). The amount of A- can be calculated from the amount of NaOH used.
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Calculate the moles of NaOH used at the equivalence point: Moles of NaOH = Molarity * Volume = 0.1 M * 40 ml = 0.1 M * 0.04 L = 0.004 moles
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Since NaOH and HA react in a 1:1 ratio, the moles of A- formed is also 0.004 moles.
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When HCl is added, it reacts with A- to form HA. The moles of HCl added can be calculated as: Moles of HCl = Molarity * Volume = 0.1 M * 20 ml = 0.1 M * 0.02 L = 0.002 moles
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The moles of A- remaining after reaction with HCl is: Moles of A- = Moles of A- initially - Moles of HCl = 0.004 moles - 0.002 moles = 0.002 moles
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The moles of HA formed from the reaction of A- with HCl is equal to the moles of HCl, which is 0.002 moles.
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The pH of the solution is given by the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) We know the pH is 5.0, and [A-] and [HA] are both 0.002 moles. Therefore, the equation simplifies to: 5.0 = pKa + log(1) Since log(1) is 0, the pKa of the acid is 5.0.
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