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....

How many mole of HCl are required to prepare 1 L of buffer solution (

1.	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» 8.85× 10^-3 is your required ANSWEREXPLANATION: NaCl + HCl is not a buffer but if HCl is LESS AMOUNT then it GIVES a buffer. NaCl + HClMole added 0.01 αMole after reaction (0.01-α) 0This is buffer of HCN+NaCNHence, pH = - log ka+ log( 0.01-a/a)8.5 = - log 4.1 × 10^-10 + log ( 0.01-a/a)a = 8.85 × 10^-3Hope it will help you ✌️