What is the PH of a buffer made with 0.18M HCN and 0.097M KCN?

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) ?

Solutions

Expert Solution

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

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

queen_honey_blossom answered 1 year ago

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