Question

Which solution would be the best choice to make a buffer with a pH of 9.2? CH3COONa / CH3COOH (Ka = 1.8 x 10–5) NH3 / NH4Cl (Ka = 5.6 x 10–10) NaOCl / HOCl (Ka = 3.2 x 10–8) NaNO2 / HNO2 (Ka = 4.5 x 10–4 )

Answer 1

For first buffer, Ka = 1.8 * 10^-5

pKa = -log Ka = -log (1.8 * 10^-5) = 4.74

for second buffer, Ka = 5.6 * 10^-10

pKa = -log(5.6 * 10^-10) = 9.25

for third buffer,Ka = 3.2*10^-8

pKa = -log( 3.2*10^-8) = 7.49

for forth buffer, Ka = 4.5 *10^-4

pKa = -log (4.5 *10^-4) = 3.35

for buffer , pH = 9.2

From Henderson Hasselbalch equation,

pH = pKa + log[salt]/[acid]

for [acid]=[salt]

pH = pKa

thus, option 2 is correct as pH is approx equal to pKa for this

Hence best choice is: NH₃/ NH₄Cl