Asked by Anonymous
Consider a solution containing 0.150 M Ba(OH)2. Calulate the pH.
Answers
Answered by
DrBob222
Ba(OH)2 ==> Ba^2+ + 2OH^-
0.150M Ba(OH)2 = 2*0.150M in OH.
pOH = -log(H^+), then pH from
pH + pOH = pKw = 14.0
0.150M Ba(OH)2 = 2*0.150M in OH.
pOH = -log(H^+), then pH from
pH + pOH = pKw = 14.0
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