Download Calculating pH of Acids & Bases: Equilibria, Buffers, & Titration and more Summaries Chemistry in PDF only on Docsity!
Acid Base Equilibria
Bronsted-Lowry Definition of Acid Base Behaviour Calculating pH Where [H+] is the concentration of hydrogen ions in the solution Calculating pH of strong acids
Finding [H+] from pH
Ionic Product for Water
A Bronsted-Lowry acid is defined as a substance that can donate a proton. A Bronsted-Lowry base is defined as a substance that can accept a proton. Each acid is linked to a conjugate base on the other
side of the equation.
HNO 3 + HNO 2 ⇌ NO 3 –^ + H 2 NO 2 +
Acid 1 Base 2 Base 1 Acid 2
HCOOH + CH 3 (CH 2 ) 2 COOH ⇌ HCOO–^ + CH 3 (CH 2 ) 2 COOH 2 +
Acid 1 Base 2 Base 1 Acid 2 In these reactions the substance with bigger Ka will act as the acid
pH = - log [H+]
Strong acids completely dissociate in water The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid. For HCl and HNO 3 the [H+(aq)] will be the same as the original concentration of the acid. For 0.1M HCl the pH will be –log[0.1] =1. Always give pH values to 2d.p. In the exam
[H+] = 10 -pH
On most calculators this is done by pressing Inv (or 2nd function) log - number(pH) Example 1 What is the concentration of HCl with a pH of 1.35? [H+] = 1 x 10-1.35^ = 0.045M In all aqueous solutions and pure water the following equilibrium occurs: H 2 O (l) H+(aq) + OH-(aq) Because [H 2 O(l)] is much bigger than the concentrations of the ions, we assume its value is constant and make a new constant Kw
Finding pH of Pure Water Calculating pH of Strong Base This equilibrium has the following equilibrium expression At 25oC the value of Kw for all aqueous solutions is 1x10-14^ mol^2 dm- The Kw expression can be used to calculate [H+(aq)] ions if we know the [OH-^ (aq)] ions and vice versa. Pure water/ neutral solutions are neutral because the [H+(aq) ] = [OH-(aq)] Using Kw = [H+ (aq) ][OH-(aq) ] then when neutral Kw = [H+(aq)]^2 and [H+(aq)] = √Kw At 25oC [H+(aq) ] = √ 1x10-14^ = 1x10-7^ so pH = 7 Example 2 : Calculate the pH of water at 50ºC given that Kw = 5.476 x 10-14^ mol^2 dm-6^ at 50ºC [H+(aq)] = √Kw = √5.476 x 10-14^ =2.34 x 10-7^ mol dm-3^ pH = - log 2.34 x 10-^7 = 6. It is still neutral though as [H+ (aq) ] = [OH- (aq)] At different temperatures to 25oC the pH of pure water changes. Le Chatelier’s principle can predict the change. The dissociation of water is endothermic so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H+^ ions and a lower pH. For bases we are normally given the concentration of the hydroxide ion.
To work out the pH we need to work out [H+(aq)] using the kw
expression.
Strong bases completely dissociate into their ions. NaOH Example 3: What is the pH of the strong base 0.1M NaOH Assume complete dissociation. Kw = [H+(aq)][OH-(aq)] = 1x10- [H+(aq)] = kw/ [OH-^ (aq)] = 1x10-14^ / 0.1 = 1x10-13M pH = - log[1x10-13] =13.
Working out pH of a weak acid at half equivalence
Diluting An Acid or Alkali Comparing The pH of a Strong Acid and A weak Acid After Dilution 10, 100 and 1000 Times Because pH is a logarithmic scale, diluting a strong acid 10 times will increase its pH by one unit, and diluting it 100 times would increase its pH by two units When a weak acid has been reacted with exactly half the neutralisation volume of alkali, the above calculation can be simplified considerably At half neutralisation we can make the assumption that [HA] = [A-] Example 7 What is the pH of the resulting solution when 25cm^3 of 0.1M NaOH is added to 50cm^3 of 0.1M CH 3 COOH (ka 1.7 x 10-5) From the volumes and concentrations spot it is half neutralisation (or calculate) pH = pka = -log (1.7 x 10-5^ ) = 4. pH of diluted strong acid pH of diluted base Example 8 Calculate the new pH when 50.0 cm^3 of 0.150 mol dm-3^ HCl is mixed with 500 cm^3 of water
Weak acids would not change in the same way as when they are diluted. They increase by less than 1 unit CH 3 CH 2 CO 2 H +H 2 O (^) ⇌ H 3 O+^ + CH 3 CH 2 CO 2 Diluting the weak acid pushes the equilibrium to the right so the degree of dissociation increases and more H+^ ions are produced meaning pH increases less than expected Buffer Solutions How Buffer Solutions Work
The buffer contains a reservoir of HA and A—^ ions
A Buffer solution is one where the pH does not change significantly if small amounts of acid or alkali are added to it An acidic buffer solution is made from a weak acid and a salt of that weak acid ( made from reacting the weak acid with a strong base). Example : ethanoic acid and sodium ethanoate CH 3 CO 2 H and CH 3 CO 2 -^ Na+ A basic buffer solution is made from a weak base and a salt of that weak base ( made from reacting the weak base with a strong acid). Example : ammonia and ammonium chloride NH 3 and NH 4 +Cl In an ethanoic acid buffer CH 3 CO 2 H (aq) (^) ⇌ CH 3 CO 2 -^ (aq) + H+^ (aq) Acid conjugate base In a buffer solution there is a much higher concentration of the salt CH 3 CO 2 -^ ion than in the pure acid i.e there is a large reservoir of the salt anions A- If small amounts of acid is added to the buffer: Then the above equilibrium will shift to the left removing nearly all the H+ ions added, CH 3 CO 2 -^ (aq) + H+(aq) ⇌ CH 3 CO 2 H (aq) As there is a large concentration of the salt ion in the buffer the ratio [CH 3 CO 2 H]/ [CH 3 CO 2 - ] stays almost constant, so the pH stays fairly constant. If small amounts of alkali is added to the buffer. The OH-^ ions will react with H+^ ions to form water. H+^ + OH H 2 O The Equilibrium will then shift to the right to produce more H+^ ions. CH 3 CO 2 H (aq) (^) ⇌ CH 3 CO 2 -^ (aq) + H+^ (aq) Some ethanoic acid molecules are changed to ethanoate ions but as there is a large concentration of the salt ion in the buffer the ratio [CH 3 CO 2 H]/ [CH 3 CO 2 - ] stays almost constant, so the pH stays fairly constant. Learn these explanations carefully and be able to write the equilibrium to illustrate your answer.
Calculating Change In pH of Buffer On Addition of Alkali Buffering Action in Blood Titration Curves Constructing a pH curve Strong Acid – Strong Base If a small amount of alkali is added to a buffer then the moles of the acid would reduce by the number of moles of alkali added and the moles of salt would increase by the same amount so a new calculation of pH can be done with the new values A carbonic acid– hydrogencarbonate equilibrium acts as a buffer in the control of blood pH Equilibrium H 2 CO 3 ⇌ H+ + HCO 3 – The H 2 CO 3 /HCO 3 –^ buffer is present in blood plasma, maintaining a pH between 7.35 and 7.45. Adding alkali reacts with H+^ so the above Equilibrium would shift right forming new H+^ and more HCO 3 – Measure initial pH of the acid Add alkali in small amounts noting the volume added Stir mixture to equalise the pH Measure and record the pH to 1 dp When approaching endpoint add in smaller volumes of alkali Add until alkali in excess Calibrate meter first by measuring known pH of a buffer solution. This is necessary because pH meters can lose accuracy on storage Can improve accuracy by maintaining constant temperature There are 4 main types of curve
- Strong acid and strong base
- Weak acid and strong base
- Strong acid and weak base
- Weak acid and weak base
Weak Acid – Strong Base e.g. CH 3 CO 2 H and NaOH The Key points to sketching a curve: Initial and final pH Volume at neutralisation General Shape (pH at neutralisation) Half neutralisation volume At ½ the neutralisation volume the [HA] = [A-] If we know the Ka we can then work out the pH at ½ V or vice versa
So Ka= [H+] and pKa = pH
Strong Acid – Weak Base Weak Acid – Weak Base e.g. CH 3 CO 2 H and NH 3
