What is the ph of 79.9 ml of a solution which is 0.66 M in NaCN...

What is the ph of 79.9 ml of a solution which is 0.66 M in NaCN and 0.55 M in HCN? for HCN use Ka=4.9 x 10^(-10)! PLEASEEE HELP!

Expert Solution

V = 79.9 ml

M = 0.66 NaCN

M = 0.55 HCn

Ka = 4.9*10^-10

pKa = -log(Ka) = -log(4.9*10^-10) = 9.31

this is a buffer, apply buffer equation

pH = pKa+ log(A-/HA)

pH = 9.31+ log(0.66 / 0.55 = 9.3891

queen_honey_blossom answered 2 years ago

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