J.R. S. answered 12/18/20
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
pH = -log [H+]
pOH = -log [OH-]
pH = pOH = 14
[H+][OH-] = 1x10-14
1). 8.3x10-2 M Ba(OH)2 ==> Ba2+ + 2OH-; [OH-] = 2 x 8.3x10-2 = 0.166 M
pOH = -log 0.166 = 0.78
pH = 14 - 0.78 = 13.2
2). 9.8x10-9 M HCN; HCN is a weak acid so doesn't ionize completely. You need to look up the Ka for HCN and then solve Ka = [H+][CN-]/[HCN] = (x)(x)/9.8x10-9 - x and solve for x = [H+]. Take -log to get pH
3). 1.7x10-3 M Ca(ClO)2 = Ca2+ + 2ClO- and [ClO-] = 2 x 1.7x10-3 = 3.4x10-3 M
Look up Kb for ClO- or Ka for HClO and calculate Kb for ClO- and then use in Kb = [HClO][OH-]/[ClO-]
(x)(x) / 3.4x10-3 - x and solve for x which will = [OH-]. Take -log to get pOH
4). Ammonium benzoate is the salt of a weak acid (benzoic acid) and a weak base (NH3). The pH will depend on the relative strength of the Ka for benzoic acid and NH4+. Look up the values and determine which one affects pH more, and that will tell you if the solution will be acidic or alkaline or neutral (neutral only if they are the same values, and not likely)