Assuming #"Ba"("OH")_2# dissociates completely, what is the #"pH"# of #"0.015 M"# #"Ba"("OH")_2# to three sig figs?

2 Answers

Answer:

12.2

Explanation:

#pOH=−log(OH^−)#

Arrhenius acid-base reaction

#Ba(OH)_2 -> Ba^"+2"+2OH^−#

#(OH^−) = 2(0.0075M) = 0.015M(OH^-)#

pOH= −log(0.015) = −(−1.82) = 1.82

pH + pOH= 14

14−1.82 = 12.2

Jul 3, 2017

#Ba(OH)_2 rightleftharpoons Ba^(+2) + 2OH^-#

Given 0.00750M #Ba(OH)_2#
=> #[OH^-] = 2(0.00750M) = 0.015M( OH^-)#

=> #pOH=-log[OH^-] = -log(0.015) = -(-1.82) = 1.82#

=> #pH = 14 - pOH = 14 - 1.82 = 12.2#