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

2 Answers

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