Question

The ionization constant of a certain weak is ${10}^{-4}.$ What be the [salt] to [acid] ratio if we have to prepare a buffer with pH = 5 using the acid and one of the salts?

• A. 1 : 10
• B. 10 : 1
• C. 5 : 4
• D. 4 : 5

Answer 1

The correct option is B 10 : 1

Given, $K_a={10}^{-4},pH=5$
According to Henderson-Hasselbalch equation,
$pH=p{K}_b+log\frac{\left[salt\right]}{\left[acid\right]}$
$pH=-log{K}_b+log\frac{\left[salt\right]}{\left[acid\right]}$
$5=-log{10}^{-4}+log\frac{\left[salt\right]}{\left[acid\right]}$
$5=4+log\frac{\left[salt\right]}{\left[acid\right]}$
$log\frac{\left[salt\right]}{\left[acid\right]}=1$
$\frac{\left[salt\right]}{\left[acid\right]}=antilog1=10$
Hence, the [salt] to [acid] ratio is 10:1