Question

Calculate the pH when 59.0 mL of 0.271 M of a certain monoprotic weak acid, HA, is mixed with 59.0 mL of 0.271 M sodium hydroxide solution at 25 °C. For HA, the Ka is 8.5× 10–5 M

Answer 1

Given:

M(HA) = 0.271 M

V(HA) = 59 mL

M(NaOH) = 0.271 M

V(NaOH) = 59 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.271 M * 59 mL = 15.989 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.271 M * 59 mL = 15.989 mmol

We have:

mol(HA) = 15.989 mmol

mol(NaOH) = 15.989 mmol

15.989 mmol of both will react to form A- and H2O

A- here is strong base

A- formed = 15.989 mmol

Volume of Solution = 59 + 59 = 118 mL

Kb of A- = Kw/Ka = 1*10^-14/8.5*10^-5 = 1.176*10^-10

concentration ofA-,c = 15.989 mmol/118 mL = 0.1355M

A- dissociates as

A- + H2O -----> HA + OH-

0.1355 0 0

0.1355-x x x

Kb = [HA][OH-]/[A-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.176*10^-10)*0.1355) = 3.993*10^-6

since c is much greater than x, our assumption is correct

so, x = 3.993*10^-6 M

[OH-] = x = 3.993*10^-6 M

use:

pOH = -log [OH-]

= -log (3.993*10^-6)

= 5.3987

use:

PH = 14 - pOH

= 14 - 5.3987

= 8.6013

Answer: 8.60