Question

A solution contains 0.306 M HA (Ka = 6.43 ⋅ 10 − 5 ) and 0.683 M NaA. What is the pH of this solution? What is the pH of this solution after 0.233 mol of HCl are added to 1.00 L of this solution? What is the pH of this solution after 0.466 mol of HCl are added to 1.00 L of this solution?

Answer 1

1)

Ka = 6.43*10^-5

pKa = - log (Ka)

= - log(6.43*10^-5)

= 4.192

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.192+ log {0.683/0.306}

= 4.54

Answer: 4.54

2)

mol of HCl added = 0.233 mol

A- will react with H+ to form HA

Before Reaction:

mol of A- = 0.683 M *1.0 L

mol of A- = 0.683 mol

mol of HA = 0.306 M *1.0 L

mol of HA = 0.306 mol

after reaction,

mol of A- = mol present initially - mol added

mol of A- = (0.683 - 0.233) mol

mol of A- = 0.45 mol

mol of HA = mol present initially + mol added

mol of HA = (0.306 + 0.233) mol

mol of HA = 0.539 mol

Ka = 6.43*10^-5

pKa = - log (Ka)

= - log(6.43*10^-5)

= 4.192

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.192+ log {0.45/0.539}

= 4.113

Answer: 4.11

3)

mol of HCl added = 0.466 mol

A- will react with H+ to form HA

Before Reaction:

mol of A- = 0.683 M *1.0 L

mol of A- = 0.683 mol

mol of HA = 0.306 M *1.0 L

mol of HA = 0.306 mol

after reaction,

mol of A- = mol present initially - mol added

mol of A- = (0.683 - 0.466) mol

mol of A- = 0.217 mol

mol of HA = mol present initially + mol added

mol of HA = (0.306 + 0.466) mol

mol of HA = 0.772 mol

Ka = 6.43*10^-5

pKa = - log (Ka)

= - log(6.43*10^-5)

= 4.192

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.192+ log {0.217/0.772}

= 3.641

Answer: 3.64