Question

How many mole of HCl are required to prepare 1 L of buffer solution ( containing NaCN+ HCl) of pH 8.5 using 0.01 g formula mass of NaCN
(K of HCN = 4.1 × 10^ )
a. 8.85× 10^-3
b.8.75× 10^-2
c . 8.85×10^-4
d. 8.85×10^-2
Answer this and I will follow u....​

Answer 1

Answer:

8.85× 10^-3 is your required answer

Explanation:

NaCl + HCl is not a buffer but if HCl is less amount then it gives a buffer.

NaCl + HCl

Mole added 0.01 α

Mole after reaction (0.01-α) 0

This is buffer of HCN+NaCN

Hence, pH = - log ka+ log( 0.01-a/a)

8.5 = - log 4.1 × 10^-10 + log ( 0.01-a/a)

a = 8.85 × 10^-3

Hope it will help you ✌️