4. A 100.0 mL sample of 0.18MHI is titrated with 0.27MKOH. Determine the pH of the solution after the addition of 110.0
Posted: Fri Jul 15, 2022 4:48 pm
4. A 100.0 mL sample of 0.18MHI is titrated with 0.27MKOH. Determine the pH of the solution after the addition of 110.0 mL of KOH. V of HCN=100=0.1 L Molarity = Volime (L) moles Mo I of HCN=0.18M moles : Molanty ×M(L) MoI of COH=0.27M Moles of HEN=θ.18×θ.1 L=θ.018moles V of KOH added =110 mL Moles of KOH=θ.27×θ.11θ=θ.9297moles Moles of HCN = moles of KOH KOH→K++OH−HCN→H++CN−