In: Chemistry
What is the PH of a buffer made with 0.18M HCN and 0.097M KCN? How does the PH change when 0.0025M of HCl is added to the buffer if the Ka (HCN) = (4.0*10^-10) ?
1)
Ka = 4*10^-10
pKa = - log (Ka)
= - log(4*10^-10)
= 9.3979
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.3979+ log {0.097/0.18}
= 9.13
Answer: 9.13
2)
Added HCl will react with CN- to form HCN
[CN-] = 0.097 - 0.0025 = 0.0945 M
[HCN] = 0.18 + 0.0025 = 0.1825 M
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.3979+ log {0.0945/0.1825}
= 9.11
Answer: 9.11