a 50.0 mL solution initially contains 0.300 M HCIO and 0.250 M KClO. what is the...

Question

Question

a 50.0 mL solution initially contains 0.300 M HCIO and 0.250 M KClO. what is the new pH after 1.0 mL of 2.0 M HBr is added ?

Ka = 2.9 x 10^-8

Answer 1

Answer #1

initially

millimoles of HClO = 50 x 0.3 = 15

millimoles of KClO = 50 x 0.25 = 12.5

millimoles of HBr added = 1.0 x 2.0 = 2.0

after HBr added

millimoles of HClO = 15 + 2.0 = 17.0

millimoles of KClO = 12.5 - 2.0 = 10.5

total volume = 50 + 1 = 51 mL

[HClO] = 17.0 / 51 = 0.33 M

[KClO] = 10.5 / 51 = 0.206 M

pH = pKa + log [KClO] / [HClO]

pKa= - log Ka = - log [2.9 x 10^-8] = 7.54

pH = 7.54 + log [0.206] / [0.33]

pH = 7.34

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Answer 2

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