Complete the following
table.
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the formulas that is use ful is
[H3O+] [OH-] = 1.0 x 10-14
pH = -log[H3O+]
pOH = -log[OH-]
pH + pOH = 14
Part A
[H3O+] = 6.4 × 10−3
[OH-] = 1.0 x 10-14 / 6.4 × 10−3 = 1.56 x 10-12
pH = -log(6.4 × 10−3) = 2.19
pOH = 14-2.19 = 11.8
acidic
Part b
[OH-] = 9.3 × 10−3
[H3O+] = 1.0 x 10-14 / 9.3 × 10−3 = 1.07 x 10-12
pH = -log[1.07 x 10-12] = 11.97
pOH = 14 - 11.97 = 2.03
Basic
Part C
pOH = 11.68
pH = 14-11.68 = 2.32
[H3O+] = 10-pH = 10-2.32 = 4.8 x 10-3
[OH-] = 10-pOH = 10-11.68 = 2.1 x 10-12
acidic
Part D
pH = 9.05
pOH = 14 - 9.05 = 4.95
[H3O+] = 10-pH = 10-9.05 = 8.9 x 10-10
[OH-] = 10-pOH = 10-4.95 = 1.12 x 10-5
basic
Part e
[OH-] = 2.3× 10−8
[H3O+] = 1.0 x 10-14 /2.3 × 10−8 = 4.34 x 10-7
pH = -log[4.34 x 10-7] = 6.36
pOH = 14 - 6.36 = 7.63
slightly acidic (or nuetral)
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