# Calculate the pH of 500ml of 1M hydrazoic acid. Ka=2.5x10^-5?

Calculating the pH isn't too tough, and I'm happy to help with any questions you may have. Just comment below. Start with the dissociation equation, HA <=> [H+][A-], meaning that Ka = [H+][A-]/[HA]. If we have 1 molar hydrazoic acid, then how much will dissociate? Well, the same amount of H+ that is created will be equal to the amount of A- that is created, since hydrazoic acid is a monoprotic acid (loses only 1 proton). We don't know what that amount is right now, but let's call it x. In other words, x = [H+] = [A-]

Ka = 2.5x10^-5 = (x)(x)/(1 molar)

x = 0.005

Since x = [H+], we know that [H+] = 0.005, and since pH = -log([H+]), pH = -log(0.005) = 2.3

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