Question

In an acidic buffer the ratio $\left[acid\right]/\left[salt\right]$ is $0.5$. If we make the concentration of acid, twice of its initial value then what is value of $10$ $\Delta pH.[\log 2=0.3]$

Answer 1

For acidic buffer,
$pH=p{K}_a+\log \frac{\left[S\right]}{\left[A\right]}\text{Given, ratio of [acid]/[salt] is 0.5}\text{Hence, , Ratio of [salt]/[acid] is 2}pH=p{K}_a+\log 2$
Now we doubled the concentration of acid. So $\frac{\left[S\right]}{\left[A\right]}=1$
$pH=p{K}_a+\log 1pH=p{K}_a\therefore \Delta pH=p{K}_a+\log 2-p{K}_a=\log 2=0.3\therefore 10\Delta pH=3$