What is the pH of a 0.0035 M Ba(OH)2 solution?
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The formula for finding the pH of 0.0035 M Ba(OH)2 is = 14.000 - (- log [OH^(-1) moles])
I assume Barium Hydroxide is a strong base. Therefore all the hydroxide ions ionize.
0.0035 * 2 = 0.007 moles of OH, because each molecule of Ba(OH)2 has 2 hydroxide ions.
pH of this solution = 14 - (-log [0.007]) = 14 - (2.1549) = 11.845 pH
Answer choice D ) 11.85
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