Question- Share with your friends Share 0 Vartika Jain answered this Dear Student, pH = pKa+log[Salt][Acid]or, 9=-log(5×10-10)+log[Salt][Acid]or, log[Salt][Acid]=9+log(5×10-10) = 1-.6990[Salt][Acid]=antilog (1.6990)=0.5Volume of 5M KCN solution required to be added to 10 ml of 2M HCN in order to maintain pH=9 isV=0.5×2M×10 ml×10005M×1000= 2 mlThe correct answer is (C) 1 View Full Answer
