Assuming $Ba {(OH)}_{2}$ $"Ba"("OH")_2$ dissociates completely, what is the $pH$ $"pH"$ of $0.015 M$ $"0.015 M"$ $Ba {(OH)}_{2}$ $"Ba"("OH")_2$ to three sig figs?

Question

9141 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-03T22:19:21

12.2

$p O H = - log (O H^{-})$ $pOH=−log(OH^−)$

Arrhenius acid-base reaction

$B a {(O H)}_{2} \to B a^{+2} + 2 O H^{-}$ $Ba(OH)_2 -> Ba^"+2"+2OH^−$

$(O H^{-}) = 2 (0.0075 M) = 0.015 M (O H^{-})$ $(OH^−) = 2(0.0075M) = 0.015M(OH^-)$

pOH= −log(0.015) = −(−1.82) = 1.82

pH + pOH= 14

14−1.82 = 12.2

Answer 2 · 2017-07-03T22:22:41

$B a {(O H)}_{2} ⇌ B a^{+ 2} + 2 O H^{-}$ $Ba(OH)_2 rightleftharpoons Ba^(+2) + 2OH^-$

Given 0.00750M $B a {(O H)}_{2}$ $Ba(OH)_2$
=> $[O H^{-}] = 2 (0.00750 M) = 0.015 M (O H^{-})$ $[OH^-] = 2(0.00750M) = 0.015M( OH^-)$

=> $p O H = - log [O H^{-}] = - log (0.015) = - (- 1.82) = 1.82$ $pOH=-log[OH^-] = -log(0.015) = -(-1.82) = 1.82$

=> $p H = 14 - p O H = 14 - 1.82 = 12.2$ $pH = 14 - pOH = 14 - 1.82 = 12.2$

Assuming Ba(OH)2"Ba"("OH")_2 dissociates completely, what is the pH"pH" of 0.015 M"0.015 M" Ba(OH)2"Ba"("OH")_2 to three sig figs?