20 mL of weak monobasic acid (HA) requires 20 mL 0.2M NaOH for complete titration. If pH of solution upon addition of 10 ml of this alkali to 25 ml of above solution of HA is 5.82, the pKa of the weak acid ____________
(Given $\log 2=0.3,\log 3=0.48$ )
 

Molarity of weak acid = 0.2 M
Now, 10 mL of 0.2 M NaOH +25 mL of 0.2M HA
Salt formed millimoles  $=\text{ 1}0\times 0.\text{2}=2$
Acid reacted in millimoles  $=\text{ 25 }\times 0.\text{2}=5$
Remaining acid left in millimoles $=\text{ 5}-\text{2 }=\text{ 3}$
$$\begin{gathered} \Rightarrow \underset{\left(acidicbuffer\right)}{pH}=pKa+\log \frac{\left[salt\right]}{\left[Acid\right]} \\ \Rightarrow 5.82=pKa+\log \frac{[2/v]}{[3/v]} \\ \Rightarrow 5.82=pKa+\log 2-\log 3 \\ \Rightarrow 5.82=pKa+0.3-0.48 \\ \Rightarrow pKa=6 \end{gathered}$$