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u11ohnotes18f2005(4)

u11ohnotes18f2005(4) - pH Calculations Recall that the...

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pH Calculations Recall that the hydronium ion (H 3 O 1+ ) is the species formed when hydrogen ion (H 1+ ) attaches to water (H 2 O). OH 1– is the hydroxide ion . For this class, in any aqueous sol’n, [ H 3 O 1+ ] [ OH 1– ] = 1 x 10 –14 ( or [ H 1+ ] [ OH 1– ] = 1 x 10 –14 ) If hydronium ion concentration = 4.5 x 10 –9 M, find hydroxide ion concentration. [ H 3 O 1+ ] [ OH 1– ] = 1 x 10 –14 M 10 x 4.5 10 x 1 ] O H [ 10 x 1 ] OH [ 9 - 14 - 1 3 14 - - 1 . 10 x y x = 2.2 x 10 –6 M = 0.0000022 M 2.2 –6 M 4# (# 0# 4# 7 y 7# 1# 8# (# 0# <# #
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Given : Find : A. [ OH 1– ] = 5.25 x 10 –6 M [ H 1+ ] 1.90 x 10 –9 M B. [ OH 1– ] = 3.8 x 10 –11 M [ H 3 O 1+ ] 2.6 x 10 –4 M C. [ H 3 O 1+ ] = 1.8 x 10 –3 M [ OH 1– ] 5.6 x 10 –12 M D. [ H 1+ ] = 7.3 x 10 –12 M [ H 3 O 1+ ] 7.3 x 10 –12 M Find the pH of each sol’n above. pH = –log [ H 3 O 1+ ] ( or pH = –log [ H 1+ ] ) A. pH = –log [ H 3 O 1+ ] = –log [1.90 x 10 –9 M ] On a graphing calculator… pH = 8.72 B. 3.6 C. 2.7 D. 11.1 A few last equations… pOH = –log [ OH 1– ] pH + pOH = 14 [ H 3 O 1+ ] = 10 –pH ( or [ H 1+ ] = 10 –pH ) [ OH 1– ] = 10 –pOH 0# ORJ# 4# 1# < 0# <# (# #
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If pH = 4.87, find [ H 3 O 1+ ].
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