pH of a Solution When Weak Acid and Strong Base are Mixed

What happens to the pH of a weak acid as you add strong base to it?

For example, aqueous sodium hydroxide solution (NaOH_(aq)) is a strong base and acetic acid (CH₃COOH_(aq)) is a weak acid.

If you add 10 mL of 0.20 mol/L NaOH_(aq) to 25 mL of 0.10 mol/L CH₃COOH_(aq), what is the pH of the resultant solution?

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IUPAC Periodic Table Challenge 2020

The IUPAC Periodic Table Challenge is back in 2020!

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Elimination of HX from Haloalkanes

One way to synthesize alkenes is to remove hydrogen halide (HX) from a haloalkane (alkyl halide) in an elimination reaction.

Learn how in AUS-e-TUTE's new tutorial Dehydrohalogenation of Haloalkanes

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Question Impossible?

This was the first question on the 2019 Western Australia ATAR Chemistry Course Examination 2019

See if you can answer it ...

Boric acid, which is a weak acid, was titrated with standardised sodium hydroxide
solution.

Which one of the indicators listed below would be the most suitable to use in this
titration?

	Indicator	Range of colour change  (pH)
(a)	thymol blue	1 – 3
(b)	bromocresol green	3.8 – 5.4
(c)	cresolphthalein	8 – 10
(d)	alizarin yellow	10 – 12

(Note: no further information was given in the question nor in the accompanying Data Book).

You've probably narrowed the answer to either (c) or (d) using a couple of common assumptions:

• these are aqueous solutions of acid and base

• the temperature of the solutions is 25C

The salt of a weak acid and a strong base in aqueous solution at 25C will have a pH greater than 7.

So, what now? We need more information, like the value of the acid dissociation constant for boric acid and some information about concentrations and volumes.

Boric acid is a weak acid, Ka = 5.8exp-10 (this information was not provided!)

Because Ka is so low, boric acid is essentially a monoprotic acid, so the salt produced is essentially NaH2BO3

Sodium ions will not hydrolyse but H2BO3- will hydrolyse.

H2BO3- + H2O --> H3BO3 + OH-

Kb for the hydrolysis of H2BO3-: 
Kb = Kw/Ka 
= 1exp-14/5.8exp-10 
= 1.7exp-5
We need to know the concentration of the salt,
 this wasn't given in the question,
 so we are going to assume a 1.0 mol/L solution 
(because the numbers are nice)
[H3BO3] = x
[OH-] = x
[H2BO3-] = 1.0 - x 
and assume x is negligible compared to 1.0
 therefore [H2BO3-] ~ 1.0

Kb = [H3BO3][OH-]/[H2BO3-]

1.7exp-5 = x2/[1.0]
take the square root of both sides:
x = 4.1exp-3 = [OH-]
pOH = -log10 [OH-] 
= -log[4.1exp-3] 
= 2.4

For aqueous solutions at 25C:
pH = 14 - pOH 
= 14 - 2.4 
= 11.6

So..... drum roll please .... the answer is (d)


Ofcourse, if the concentration of salt was 1exp-5 mol/L, 
the pH of the solution would be 9.1 .... 
and then the answer would be (c)

If you were unfortunate enough to sit this exam...
 then the answer according to the examiners report was (c) ...
but no details of how they arrived at this answer are provided.