Ka = 1.2*10^-2
pKa = - log (Ka)
= - log(1.2*10^-2)
= 1.921
use:
pH = pKa + log {[conjugate base]/[acid]}
= 1.921+ log {0.5/0.5}
= 1.921
Buffer range is 1.92-1 to 1.92+1 that is 0.92 to 2.92
Answer: option 1, 3
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