1)
Ka = 3.5*10^-8
pKa = - log (Ka)
= - log(3.5*10^-8)
= 7.456
use:
pH = pKa + log {[conjugate base]/[acid]}
= 7.456+ log {0.368/0.368}
= 7.456
Answer: 7.46
2)
NaOH is strong base and donates OH-.
This OH- will react with acidic component of buffer which is HClO
The net ionic reaction is:
HClO + OH- -> ClO- + H2O
A buffer solution is made that is 0.368 M in HCIO and 0.368 M in NaClO....
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