Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.
Asked by maham237 @ in Chemistry viewed by 383 People
Asked by maham237 @ in Chemistry viewed by 383 People
Answered by maham237 @
Given:
Concentration of Ba(OH)2 = 0.348 M
To determine:
pOH of the above solution
Explanation:
Based on the stoichiometry-
1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion
Therefore, concentration of OH- ion = 2*0.348 = 0.696 M
pOH = -log[OH-] = - log[0.696] = 0.157
Ans: pOH of 0.348M Ba(OH)2 is 0.157
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