Question

What is the pH of a solution that has [HCN]=0.25M and [NaCN]=0.50M ? (ka of HCN=4.9 x 10^-10)

Answer 1

use the Henderson-Hasslebalch equation: \(pH=pKa+log\dfrac{[CN^-]}{[HCN]}\)

Answer 2

@aaronq thank you so much ! For this problem first I need to find pKa ? or can I just plug in my given values ?

Answer 3

no problem! yeah find the the pKa is just the -log of the ka; pKa=-log(Ka)

then plug your values in.

Answer 4

Got it thank you so much !

Answer 5

good stuff, no problem !