Correct Answer - `K_(b) = 1.5 xx 10^(-9)`
pH = 3.44
We know that, `pH=-log[H^(+)]`
`therefore[H^(+)]=3.63xx10^(-4)`
Then, `K_(h)=((3.63xx10^(-4))^(2))/(0.02)" "(because"concentration = 0.02M")`
`rArrK_(h)=6.6xx10^(-6)`
Now, `K_(h)=(K_(w))/(K_(a))`
`rArrK_(a)=(K_(w))/(K_(h))=(10^(-14))/(6.6xx10^(-6))`
`=1.51xx10^(-9)`