Acids and Bases
How do you calculate the pH of a diluted acid?
If the volume increases by X then the concentration will decrease by X, as they are inveserely proportional.
What is the equation for Ka?
Ka = 10^-pKa
What is Ka?
Ka is the acid dissociation constant - measures strength of the acid. The higher the Ka, the stronger the acid.
What is the value of Kw at room temperature?
Kw = 1.00 x 10^-14 mol2dm-6 If Kw is not given assume this value.
What are the rules of pH?
pH values are given to 2dp and have no units, while [H+] are given to 3sf and have the units moldm-3. - To dilute a solution from a pH 1 to a pH 4 (just 3 pH units) would require dilution by 10 x 10 x 10 = 10^3. - A solution with a pH of 1 contains 10^13 times more H+(aq) ions as a solution with a pH of 14.
What is the formula for pH?
pH=-log10[H+]
What is pKa?
pKa = - log10Ka pKa values are much more manageable than Ka and it is much easier to compare relative acidic strengths using pKa values than Ka values.
How do you calculate the pH of a strong base?
~For bases we are normally given the concentration of the hydroxide ion, which is [OH-] ~By Kw, we know that [H+] = Kw / [OH-] ~ pH = - log (Kw / [OH-] )
50cm3 of 0.100 moldm-3 hydrochloric acid is diluted to 100cm3 with water. What is the change in pH?
(1) Find the concentration of diluted HCl. The diluting to 100cm3, the volume has increased by X2. Therefore, the concentration has been halved to 0.0500 moldm-3. This is because concentration and volume are indirectly proportional. So [diluted HCl] = 0.0500moldm-3 (2) Find the pH values before and after dilution. HCl is a strong monobasic acid and completely dissociates. [H+(aq)] = [HCl(aq)] Before dilution, [H+] = 0.100 moldm-3 pH = - log[0.100] = 1.00 After dilution, [H+] = 0.0500 moldm-3 pH = - log[0.0500] = 1.30 pH change = 1.30 - 1.00 = 0.30
How can Ka for a weak acid be determined experimentally?
(1) Prepare a standard solution of the weak acid of known concentration. (2) Measure the pH of the standard solution using a pH meter. (3) The Ka value of a weak acid can then be calculated from the acid concentration and the pH measurement.
What are the limitations of using approximations to Ka related calculations for stronger weak acids? When might the circumstances under which Ka approximates break down?
(1) The assumption that [H+] = [A-] fails for high pH conditions (pH>6), because the dissociation of water is significant resulting in an increased H+. This approximation breaks down for very weak acids or very dilute solutions. (2) The second equilibrium assumes that the concentration of acid is much greater than the H+ concentration at equilibrium, which may not be true for stronger weak acids. [HA(aq) ]start >> [H+(aq) ]eqm [HA(aq)] eqm = [H+(aq)]start - [H+(aq)]eqm approximates to [HA]eqm = [HA] start This approximation will hold for weak acids with small Ka values. It breaks down when [H+(aq)] becomes significant and there is a real difference between [HA(aq)] eqm and [HA]start - [H+(aq)] eqm. This approximation is justified for stronger weak acids with Ka > 10^-2 moldm-3 Nd for very dilute solutions.
The pKa for phenol is about 10 while the pKa value got ethanol ie about 16. Explain the relative magnitudes of these values.
-Phenol is more acidic (stronger acid) than ethanol. - The equilibrium of phenol lies further to the right. - Phenol dissociates into ions more readily in solution. - The reduction of charge density stabilises anion.
Predict how the relative values of Ka and pKa would change for each dissociation.
From 1st to 2nd to 3rd dissociations of H3P04, Ka decreases and pKa increases. The acid becomes weaker with each successive dissociation.
How is [H+] related to pH?
A low value of [H+(aq)] matches a high value of pH. A high value of [H+(aq)], matches a low pH. A change in one pH number is equal to a 10 times difference in the [H+(aq)]. For example, a pH of 1 has 10 times the concentration of H+ions as a solution with a pH of 2.
When is a solution acidic, neutral or alkaline?
A solution is acidic when: [H+] > [OH-] A solution is neutral when [H+(aq)] = [OH(aq) -] A solution is alkaline when: [H+(aq) ] < [OH-(aq) ]
How can an aqueous solution of an acid contain hydroxide ions?
An aqueous solution of an acid can contain hydroxide ions because water dissociates.
Hydrobromic acid is stronger than hydrochloric acid. Explain what this means in terms of hydrogen ions and pH.
HBr will dissociate more in solution than HCl and relaseses more H+ ions, so the concentration of hydrogen ions will be higher, so the pH will be lower.
What is the effect of concentration of Kw?
Changing the concentration of [H+] or [OH-] in solution has no effect on the value of Kw as the equilibrium will shift, changing the concentration of the other substances to keep the value of Kw the same.
What is the effect of temperature on Kw.
Changing the temperature of the solution changes the value of Kw, the hydrogen ion and the hydroxide ion concentration and the pH. All equilibrium constants are temperature dependent (and this one is no exception), so at temperatures different to 25°C the pH of pure water changes. Le Chatelier's principle can predict the change.
How are [H+] and [OH-] related in an aqueous solution?
For aqueous solutions, the product of the hydrogen-ion concentration and the hydroxide-ion concentration equals 1.0 x 10^-14
How do you calculate the pH of a strong base?
For bases we are normally given the concentration of the hydroxide ion. To work out the pH we need to work [H+] using the Kw expression.
What is a monoacidic base?
NaOH and KOH are monoacidic bases as each mole of base relases one mole of OH- ions in aqueous solution. Therefore the concentration of OH- ions is the same as the concentration of the base. A diacidic base is a base that dissociates in aqueous solutions to produce two OH- ions per molecule of that base. Examples of dibacidic bases include: Ca(OH)2, Ba(OH)2 and Mg(OH)2. A triacidic base is a base that dissociates in aqueous solutions to produce there OH- ions per molecule of that base. Examples of triacidic bases include Al(OH)3.
What is pH?
PH is a measure of the hydrogen ion concentration in a solution. pH = - log10[H+] The lower the pH, the higher the concentration of hydoegn ions in the solution.
How do you find the pH of pure water?
Pure water/ neutral solutions are neutral because the [H+] = [OH-]. For pure water, there's a 1:1 ratio of H+ and OH- ions due to dissociation. Therefore, [H+] = [OH-] and so Kw = [H+]^2. Using Kw = [H+][OH-] then when neutral Kw = [H+]^2 and [H+] = √ Kw At 25 oC [H+] = √ 1x10-14 = 1x10-7 so pH = 7
What is the equation of [H+] involving Ka and [HA].
The Ka equation can then be rearranged to find [H+] to give pH.
Explain the effect of increasing the temperature on Kw.
The dissociation of water molecules into ions is an endothermic process (because bonds are broken). According to Le Chatelier's principle, if the temperature is increased, the position of equibilibrim will shift to lower the temperature by absorbing heat. So if the temperature is increased, the position of equibilibrim shifts further to the right and moves in the endothermic, forward direction to reduce the temperature by absorbing heat. Therefore, more hydrogen ions and hydroxide ions will be formed. As the values of the hydrogen ion concentration and the hydroxide ion concentration increases, Kw increases as they are proportional (being multiplied). As pH is a measure of hydrogen ion concentration (pH = -log10[H+]) then as the temperature increases the pH gets lower - i.e. water becomes more acidic. However, pure water remains neutral even if its pH changes because [H+] = [OH-]
What is Kw?
The ionic product of water Kw = [H+] [OH-] Kw is 1.00 x 10^-14 at room temperature and pressure.
Deduce whether the ionisation of water is exothermic or endothermic.
The ionisation of water is endothermic because bonds are broken and Kw increases with temperature.
How are Ka and pKa interpreted?
The larger the Ka value, and the smaller the PKa value, the stronger the acid (dissociates more). The smaller the Ka value and the larger the pKa value, the weaker the acid (dissociates more). The larger the value of Ka, the further the position of equibilibrim is to the right (acid dissociates more).
What are the units for Ka?
The units for Ka are always moldm-3. You can show this by cancelling the units in the Ka expression.
What are the approximations for weak acid calculations?
To make the calculation easier two assumptions are made to simplify the Ka expression: (1) As the dissociation of weak acids, HA, is small you can assume that the initial concentration of the undissociated acid has remained constant (it is assumed that the concentration of the acid before dissociation is equal to the concentration of the acid at equilibrium. So [HA(aq)] eqm = [HA(aq)] initial (2) [H+(aq)] eqm = [A-(aq)] eqm because they have dissociated according to a 1:1 ratio. It assumes that the dissociation of H20 is negligible.
What happens when water dissociates?
Water can act as both an acid and a base (one water molecule acting as a base can accept a hydrogen ion from a second one acting as an acid). A hydroxomium ion and a hydroxide ion are formed when water dissociates. The hydroxomium ion is a very strong acid, and the hydroxide ion is a very string base, so they react to form water again as soon as they are formed.
Ethanoic acid contains both hydrogen ions and hydroxide ions. How can an aqueous solution of an acid contain hydroxide ions?
Water dissociates H20 ⇌ H+ + OH- 2H20 ⇌ H30+ + OH-
Why is water not included in the Kw expression?
Water is not included in the Kw expression because do little of water is ionised at any one time, that its concentration remains unchanged. Water is a constant.
How do you find the pH of a weak acid?
Weak acids only partially dissociate in aqueous solution, so the [H+] isn't equal to the acid concentration. So to find the pH of a weak acid, you need to use an equilibrium constant called the acid dissociation constant, Ka. The units of Ka are moldm-3.
Explain why the units for Kw are mol2dm-6.
[H+] and [OH-] are being multiplied together - two concentrations are being multiplied.
