22)
Ka = 3.5*10^-4
pKa = - log (Ka)
= - log(3.5*10^-4)
= 3.456
use:
pH = pKa + log {[conjugate base]/[acid]}
pH = pKa + log {[CN-]/[HCN]}
= 3.456+ log {1.5*10^-2/1.1*10^-2}
= 3.591
Answer: 3.59
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