Question

Which of the following acids (listed with Ka values) and their conjugate base would form a buffer with a pH of 2.34? a. HF, Ka = 3.5 x 10-4 b. C6H5COOH, Ka = 6.5 x 10-5 c. HClO2, Ka = 1.1 x 10-2 d. HClO, Ka = 2.9 x 10-8 e. HIO3, Ka = 1.7 x 10-1

Answer 1

Answer:

HClO2

Explanation:

a. HF

$K_a=3.5\times10^{-4}\\pKa=-logKa\\\;\;\;= -log(3.5\times10^{-4})=3.455$

b. C6H5COOH

$K_a=6.5\times10^{-5}\\pKa=-logKa\\\;\;\;= -log(6.5\times10^{-5})=4.1870$

c. HClO2

$K_a=1.1\times10^{-2}\\pKa=-logKa\\\;\;\;= -log(1.1\times10^{-2})=1.96$

d. HClO

$K_a=2.9\times10^{-8}\\pKa=-logKa\\\;\;\;= -log(2.9\times10^{-8})=7.53$

e. HIO3

$K_a=1.7\times10^{-1}\\pKa=-logKa\\\;\;\;= -log(1.7\times10^{-1})=0.7695$

only pKa of HClO2 is closest to the target pH of 2.34 and therefore, suitable  for the making buffer with pH 2.34.