③ Ka= 4,9x1010 HCN pason Ž H+ + ON sater At what pH is EEN J...

Question

Question

science chemistry

Answer 1

Answer #1

Using Henderson Hasselbalch equation

pH = pKa + log[CN-]/[HCN]

given [CN-]/[HCN] = 10000

Ka = 4.9 * 10^(-10)

so pKa= -logKa= 9.30

putting the values in equation we get,

pH = 9.30 + log(10000)

pH = 13.3

Answer 2

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