The equivalent point in titration of `40.0` ml of a solution of a weak monoprotic acid occurs when `35.0` of a `0.10` M NaOH solution has been added . The pH of the solution is `5.75` after the addition of `20.0` ml of NaOH solution . What is the dissociation constant of the acid ?

Explanation

To find the dissociation constant (Ka) of the weak monoprotic acid, we can follow these steps:

### Step 1: Calculate the moles of NaOH added at the equivalence point
At the equivalence point, the moles of NaOH added is given by:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.10 \, \text{mol/L} \times 35.0 \, \text{mL} = 0.10 \, \text{mol/L} \times 0.035 \, \text{L} = 0.0035 \, \text{mol} \]
This means that 0.0035 moles of the weak acid (HA) were present initially.
...