Question

18. Calculate the pH of a 0.027 M Ba(OH)2 solution. pH = Enter your answer in the provided box. How much NaOH (in grams) is needed to prepare 681 mL of solution with a pH of 9.570? g NaOH

Answer 1

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Ba(OH)₂ is strong base dissociate completly as given below

Ba(OH)₂   Ba²⁺  + 2OH^-

According to dissociation reaction 1 mole of Ba(OH)₂ produce 2 mole of OH^- ion therefore 0.027 M soluton of Ba(OH)₂ produce 0.054 M of OH^- ion solution

pOH = -log[OH^-] = -log(0.054) = 1.27

pH = 14 - pOH = 14 - 1.27 = 12.73

pH = 12.73

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pOH = 14 - pH = 14 - 9.570 = 4.43

[OH^-] = 10^-pOH = 10^-4.43 = 3.72 10^-5 M

NaOH is monobesic strong base dissociate completly therefore [OH^-] = [NaOH] = 3.72 10^-5 M

no. of mole = molarity volume of solution in liter

no. of mole of NaOH = 3.72 10^-5 0.681 = 2.53 10^-5 mole

molar mass of NaOH = 39.997 gm/mol then 2.53 10^-5 mole = 2.53 10^-5 39.997 = 0.001012 gm

0.001012 gm of NaOH required.