Acids, bases and buffers Flashcards
Definition of a Bronsted-Lowry acid.
Proton donor.
Definition of a Bronsted-Lowry base.
Proton acceptor.
Definition of a strong acid.
A proton donor that fully dissociates into its ions in solution.
Definition of a weak acid.
A proton donor that partially dissociates into its ions in a solution.
What do the terms monobasic, dibasic and tribasic refer to?
The total number of hydrogen ions in the acid that can be replaced per molecule in an acid-base reaction.
Definition of a conjugate acid-base pair.
A set of two species that transform into each other by loss or gain of a proton, i.e they consist of the same species apart from the H+.
What is the equation for find the pH?
pH = -log10[H+]
What is the equation to find [H+]?
10^-pH
Write a Kc expression for the dissociation of water.
*Refer to ABB notes.
What is the value of Kw at 25 degrees?
1 x 10^-14 mol^2dm^-6.
Calculate the pH of 0.11moldm^-3 of HCl(aq).
pH = 0.959
Calculate the pH of 2moldm^-3 HNO3(aq).
pH = -0.301
Calculate the pH of 0.5moldm^-3 of HCl(aq).
pH = 0.301
Cacluate the pH of 0.21moldm^-3 of H2SO4(aq).
pH = 0.377
Calculate the pH of 0.01moldm^-3 of H3PO4(aq).
pH = 1.52
Find [H+] in a solution of HCl(aq) of pH 2.05.
[H+] = 8.91 x 10^-3 moldm^-3
Find [H+] in a solution of CH3CH3COOH(aq) of pH 3.40.
[H+] = 3.98 x 10^-4 moldm^-3
Find [H+] in NaOH9aq) with a pH of 13.5.
[H+] = 3.16 x 10^-14 moldm^03
Find the pH of water at 298K given that Kw = 1 x10^-14 mol^2dm^-6 at 298K.
pH = 7.
Find the pH of water at 398K given that Kw = 2.5 x 10^-14 mol^2dm^-6.
pH= 6.80
Calculate the pH of 1moldm^-3 NaOH(aq).
pH = 14
Calculate the pH of 0.03moldm^-3 of KOH(aq).
pH = 12.5
Calculate the pH of 0.025moldm^-3 CaOH2(aq).
12.7
What is the equation to calculate pKa?
pKa = -log10Ka
What is the equation to calculate Ka?
Ka = 10^-pKa
How does a larger Ka value affect the equilibrium?
The larger the numerical value of Ka, the more the equilibrium lies to the right, the greater the concentration of the dissociated ions H+ and A- (numerator), and the stronger the acid.
How does a smaller Ka value affect equilibrium?
The smaller the numerical value of Ka, the more the eqilibrium is to the left, the smaller the concentration of the dissociated ions H+ and A- (nuerator, and the weaker the acid.
Why is [H+] = [A-] only an approximation when calculating the pH of weak monobasic acids using Ka?
Water also dissociates forming a very small concentration of H+ but this is so small it can be considered to be negligible compared with the [H+] from the acid.
When will the approximation: [HA]eqm = [HA]start (undissociated), be true when calculating the pH of a weak monobasic acid using Ka?
[HA]eqm = [HA]start - [H+]eqm. As the dissociation of weak acids is small, can assume [H+]eqm is very small so [HA]»_space; [H+]eqm and the decrease in the concentration of HA from the dissociation is negligible.
What is the simplification of the Ka expression when using the approximations to calculate the pH of a weak monobasic acid?
*Refer to ABB notes.
Find the pH of a weak acid; HA, which has a concentration of 0.1moldm^-3 and a Ka of 1.7 x 10^-5 moldm^-3.
2.88
Explain the limitation of the approximation: [H+] = [A-] when calculating the pH of a weak monobasic acid.
This assumes that the [H+] from water is negligible. However, at 25 degrees, the [H+] from the dissociation of water is 1 x 10^-7. For a very weak acid with a pH>6, the [H+] from the dissociation of water will be significant compared with the dissociation of the weak acid. For example: pH = 6.1, the [H+] = 10^-6.1= 7.94 x 10-5 moldm^-3.
Explain the limitation of the approximation: [HA]eqm = [HA] start (undissociated) when calculating the pH of a monobasic weak acid.
This assumes [HA]»_space; [H+]. This is not true when [H+] becomes significant. As the acid has dissociated more (stronger acid). Dissociation of water is significant. There is now a significant difference between [HA]eqm and [HA]start. So, the approximation is not valid for a stronger weak acid and for very dilute solutions.
Definition of a buffer
A system that resists change in pH when small amounts of acid or base are added.
What does a weak acid buffer solution contain?
A weak acid and the salt of a weak acid, for example: CH3COOH and CH3COO- 9ethanoic acid and ethanoate).
Describe two methods of making a weak acid buffer solution.
- The weak acid and one of its salts. For example: CH3COOH and CH3COO-.
- Adding an alkali and an excess of the weak acid. For example: NaOH and CH3COOH.
How does the buffer work on addition of acid? Use CH3COOH/CH3COO- as the conjugate base pair.
H+ from the acid reacts with CH3COO-.
CH3COO- + H+ …. CH3COOH
Equlibrium shifts to the left.
H+ ions being removed.
How does the buffer work on addition of an alkali? Use CH3COOH/CH3COO- as the conjugate base pair.
Added OH- reacts with H+.
H+ + OH- ….H2O
Equilibrium shifts to the right.
CH3COOH …. CH3COO- + H+
H+ ions are being replaced.
What is the Henderson-Hasselbach equation?
*Refer to ABB notes
Calculate the pH of a buffer solution made from 0.60moldm^-3 propanoic acid and 0.80moldm^-3 moldm^-3 sodium propanoate.
pH = 3.99
The pH of a buffer is desired to be 3.76. What concentration of methanoic acid must be mixed with 0.5moldm^-3 sodium methanoate to achieve this? Ka of methanoic acid is 1.78 x 10^-4.
[HCOOH] = 0.489 moldm^-3
Write the reaction equation for water and carbon dioxide.
CO2 + H20 …/… H2CO3
Write the dissociation equation for carbonic acid.
H2CO3 …/… H+ + HCO3-
Describe the equilibrium of blood on addition of acid.
H+ from the acid reacts with HCO3-.
H+ + HCO3- … H2CO3
Equlibrium shifts to the left.
H+ ions are removed.
Decribe the equlibrium of blood on addition of alkali.
OH- ions react with acid (the H+ of H2CO3).
H+ + OH- …. H2O
Equlibrium shifts to the right.
Realces the H+ ions.
Definition of equivalnce point
The volume of one solution that exactly reacts with the volume of the other solution.
What are the three key points for drawing pH curves?
- pH at the start.
- The volume added to reach the end point / neutralisation (the titre).
- pH at the end.
Draw a titration curve for the reaction between a strong acid, e.g. HCl, and a strong base, e.g. NaOH.
*Refer to ABB notes.
Draw a titration curve for the reaction between a weak acid and a strong base.
*Refer to ABB notes.
Draw a titration curve for the reaction between a strong acid and a weak base.
*Refer to ABB notes.
