Question

Be sure to answer all parts. An aqueous solution of a strong base has pH 10.88 at 25°C. Calculate the concentration of base in the solution: (a) if the base is LiOH. [LiOH] = M (b) if the base is Ba(OH)2. [Ba(OH)2] = × 10 M (Enter your answer in scientific notation.)

Answer 1

a) The pH of LiOH solution =10.88 pOH = 14.00 -PH = 14.00 - 10.88 = 3.12 POH = -log[OH-] [OH-] = 10 DOH = 10-312 = 0.00076 M - - - - - - Equation representing the dissociation of LiOH: LiOH →→ Lit + OH Since 1 mol of LiOH forms 1 molels of OH upon dissociation [OH-] = [LiOH] [LiOH] = [OH-] = 0.00076 M 7.6x10* M - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - b) The pH of Ba(OH), solution =10.88 pOH = 14.00 -PH = 14.00 -10.88 = 3.12 pOH = -log[OH-] [OH-] = 10 POH = 10-312 = 0.00076 M 7.6x10* M ----------- Equation representing the dissociation of Ba(OH), Ba(OH), (aq) →→ Ba* (aq) + 2OH- (aq) Since 1 mol of Ba(OH), forms 2 moles of OH upon dissociation [OH-] = 2 [Ba(OH)] [Ba(OH)2] = } [or] = *(76x104 M) = 3.8*10*M