0.1276 of a monoprotic acid (molar mass 110g/mol) was dissolved in 25ml of water.The pH of the solution was found to be 5.87 after the addition of 10.0ml of 0.0633 M NaOH.The Ka of acid is.......................
HA + NaOH ==> NaA + H2O
Calculate moles HA from the data given. Calculate moles NaOH, determine moles NaA formed and moles HA remaining, plug into Henderson-Hasselbalch equation and solve for pKa, then Ka. Post your work if you get stuck.
To find the Ka (acid dissociation constant) of the monoprotic acid, we need to determine the concentration of the acid before the addition of the NaOH.
First, let's calculate the number of moles of the acid that were dissolved in the solution:
Number of moles of acid = (mass of acid) / (molar mass of acid)
Number of moles of acid = (0.1276 g) / (110 g/mol)
Number of moles of acid = 0.00116 mol
Next, let's calculate the volume of the acid solution before the addition of NaOH:
Volume of acid solution = volume of water used
Volume of acid solution = 25 mL = 0.025 L
Now, using the volume and concentration of the NaOH added, we can find the number of moles of NaOH that reacted with the acid:
Number of moles of NaOH = (volume of NaOH) × (concentration of NaOH)
Number of moles of NaOH = (10.0 mL) × (0.0633 mol/L)
Number of moles of NaOH = 0.000633 mol
Since the acid and base react in a 1:1 ratio, the number of moles of acid that reacted is equal to the number of moles of NaOH that reacted.
Now, let's calculate the concentration of the acid after the addition of NaOH:
Concentration of acid = (number of moles of acid - number of moles of acid reacted) / volume of acid solution
Concentration of acid = (0.00116 mol - 0.000633 mol) / 0.025 L
Concentration of acid = 0.000527 mol / 0.025 L
Concentration of acid = 0.0211 mol/L
Finally, we can calculate the Ka of the acid using the concentration of the acid and the pH of the solution:
pH = -log[H+]
[H+] = 10^(-pH)
From the pH value of 5.87, we can calculate the concentration of H+ ions in the solution:
[H+] = 10^(-5.87)
[H+] = 1.41 × 10^(-6) mol/L
Since the acid is monoprotic, it can be assumed that [H+] = [A-] (concentration of conjugate base).
The concentration of the acid is given by [A-] = Ka × [HA]:
(0.0211 mol/L) × (1.41 × 10^(-6) mol/L) = Ka × (0.0211 mol/L)
Now, we can solve for Ka:
Ka = [A-] / [HA]
Ka = (0.0211 mol/L) × (1.41 × 10^(-6) mol/L) / (0.0211 mol/L)
Ka = 1.41 × 10^(-8) mol/L
Therefore, the Ka of the monoprotic acid is 1.41 × 10^(-8) mol/L.