Acid Base Equlibria

• Brønsted-Lowry Acid - these are protons (H+) donors in aqueous solution.

  e.g. HCl (aq) - H+ (aq) + Cl- (aq)
• Brønsted-Lowry Base - proton (H+) acceptor in aqueous solution.

  e.g. OH- (aq) + H+ (aq) - H2O (l)
• When a Brønsted-Lowry acid reacts with a Brønsted-Lowry base, in a reversible reaction, the products must also act as conjugate Brønsted-Lowry acids and Brønsted-Lowry bases.

  
  

  e.g. NH3 (aq) (B1) + H2O (l) (A2) - NH4+ (aq) (A1) + OH- (aq) (B2)

  
  

  in this example, NH3 accepts a proton, in forming NH4+, thus NH3 is the Brønsted-Lowry base. When viewed in reverse, NH4+ donates a proton in forming NH3, making NH4+ the conjugate acid.

  H2O donates a proton in forming OH-, making H2O the Brønsted-Lowry acid. When viewed in reverse, OH- accepts a proton in forming H2O, making OH- the conjugate base.
• PH is calculated through the concentration of hydrogen ions in solution (unitless)

  this logarithmic scale is used because of the wide range in [H+] values.

  PH =$-\log_{10}\left[H^{+}\right]$
• $\left[H^{+}\right]$ in mol/dm^3 can equally be calculated from PH

  $\left[H^{+}\right]$ = $10^{-PH}$
• Since PH is a logarithmic scale...

  dilution (of H+ ions) by x 10 will result in PH increase of 1

  dilution (of H+ ions) by x 100 will result in PH increase of 2

  dilution (of H+ ions) by x 1000 will result in PH increase of 3

  e.g.

  PH = $-\log_{10}\left(0.10\right)=$$1$

  PH = $-\log_{10}\left(0.01\right)=$$2$
• PH of strong acids - strong acids completely dissociate (shown by a one way reaction arrow: $HCl(aq)\rightarrow H^+$$(aq)+$$Cl^-(aq)$

  this means that: $\left[H^+(aq)\right]$ = [strong acid]

  
  

  with some diprotic acids (H2SO4), $\left[H^+\right]$ is no longer equal to the [strong acid]

  e.g. $H_2SO_4(aq)\rightarrow2H^+$$(aq)+$$SO_4^{2-}(aq)$

  in this case: $\left[H^+\right]=$$2\times$$\left[H_2SO_4\right]$

  (this isnt actually true, as the second $H^+$released is far weaker than the first.
• dissociation of water - water weakly dissociates

  $H_2O(l)\leftrightarrow H^+$$(aq)+$$OH^-(aq)$

  since its a reversible reaction, the equilibrium constant (Kc) can be calculated: $Kc=$$\left[\frac{\left[H^+(aq)\right]\left[OH^-(aq)\right]}{H_2O(l)}\right]$

  this expression can be rearranged to form a new constant, the ionic product of water:

  $Kc\left[H_2O(l)\right]=$$\left[H^+(aq)\right]\left[OH^-(aq)\right]$

  $Kw=$$\left[H^+(aq)\right]\left[OH^-(aq)\right]$

  $Kw=$$\left[H^+(aq)\right]^{2}$ --- only in pure water when [H+] and [OH-] are the same
• $pKw=$$-\log_{10}Kw$

  $Kw=$$10^{-pKw}$

  the smaller the pKw value, the greater the value of Kw.
• At 25 degrees, $Kw\ =$$\ 1\times10^{-14}mol^2dm^{-6}$

  the value of Kw (degree of dissociation) changes with temp, as the dissociation of water involves bond breaking - an endothermic process
• An increase in temperature shifts the equilibrium position to the right (of the water dissociation equation) to decrease the temp. This leads to an increase in the value of Kw

  At 50 degrees $Kw\ =$$\ 5.5\ \times10^{-14}mol^2dm^{-6}$
• An decrease in temperature shifts the equilibrium position to the left(of the water dissociation equation) to increase the temp. This leads to an decrease in the value of Kw.

  At 10 degrees, $Kw\ =$$\ 2.9\ \times10^{-15}mol^2dm^{-6}$
• As we know, since water dissociates in a 1:1 ratio, its PH can be calculated with the following variation of the ionic product of water $Kw\ =$$\left[H^+(aq)\right]^2$

  since the degree of dissociation varies with temp, so must the PH of water. However, no matter the value of the PH for pure water, it is still considered to be neutral as $\left[H^+(aq)\right]=$$\left[OH^-(aq)\right]$
• like acids, strong bases will completely dissociate:

  $NaOH\ (aq)\ \rightarrow\ Na^+$$(aq)+$$OH^-(aq)$

  thus $\left[OH^-(aq)\right]=$$\left[strong\ base\right]$
• to calculate the PH of strong bases, $\left[H^+\right]$ must be calculated, and you can use the expression for Kw to do this.

  $Kw\ =$$\left[OH^-(aq)\right]\left[H^+(aq)\right]$

  $\left[H^+(aq)\right]=$$\frac{Kw}{\left[OH^-(aq)\right]}$
• some bases as strong dibasic bases $\left(Ba\left(OH\right)2(aq)\right)$

  e.g. $Ba\left(OH\right)2\ (aq)\ \rightarrow\ Ba^{2+}(aq)+$$2\ OH^-(aq)$

  thus $\left[OH^-(aq)\right]=$$\ 2\ \times\left[Ba\left(OH\right)2(aq)\right]$
• weak acids only slightly dissociate in water, producing fewer hydrogen ions. This is shown by the reversible reaction symbol.

  $HA(aq)\ \leftrightarrow\ H^+$$(aq)+$$A^-(aq)$
• the equilibrium dissociation gives an expression for weak acids, where Ka = the acid dissociation constant.

  $Ka\ =$$\ \frac{\left[H^+(aq)\right]\left[A^-(aq)\right]}{\left[HA(aq)\right]}$

  the greater the value of $Ka$, the further the dissociation equilibrium lies to the right-hand side. This means acids with a greater value of Ka are stronger than those with a smaller value.
• $pKa=$$-\log_{10}\left(Ka\right)$

  $Ka\ =$$\ 10^{-pka}$
• to calculate $\left[H^+\right]$ for a weak acid, you'd use the expression for Ka, which can be simplified under the following assumptions:

  $Ka\ =$$\frac{\left[H^+(aq)\right]\left[A^-(aq)\right]}{\left[HA(aq)\right]}$

  1. $\left[H^+(aq)\right]e\ =$$\left[A^-(aq)\right]e$, as the weak acid dissociates in a 1:1 ratio.
  2. $\left[HA(aq)\right]e\ \approx\left[HA(aq)\right]i$ as the degree of dissociation of the weak acid is so small
  3. The [H+] from water is negligible
• as a result of the mentioned assumptions, $\left[H^+(aq)\right]$ for a weak acid can be calculated using the following simplified equations. (you would want the HA concentration to be at equilibrium, but since its 'so similar' to the initial concentration, you can just use that value).

  $Ka\ =$$\frac{\left[H^+(aq)\right]^2e}{\left[HA(aq)i\right]}$
• since the degree of dissociation determines the value of Ka (it actually matters since Ka is the acid dissociation constant), we can no longer assume $\left[HA(aq)\right]e\approx\left[HA(aq)\right]i$, in a Ka calculation.

  if the value of $\left[H^+(aq)\right]$ is very low, $\left[HA(aq)\right]e\approx\left[HA(aq)\right]i$, and Ka can be used via the original calculation.

  $Ka\ =$$\ \frac{\left[H^+(aq)\right]^2e}{\left[HA\ (aq)\right]i}$
• if the value of $\left[H^+(aq)\right]$ is significant, the assumption also cannot be made. This is most common when dissolving a solid weak acid into a solution. If $\left[H^+(aq)\right]$ = $x\ mol\ dm^{-3}$, $\left[HA\right]$ must have decreased by $x\ mol\ dm^{-3}$

  $Ka\ =$$\ \frac{\left[H^+(aq)\right]^2e}{\left[HA(aq)\right]i-\left[H^+(aq)\right]e}$
• the equivalence point is the middle of the vertical section of a titration curve. This is the point of exact neutralisation where THE MOLES OF ACID = THE MOLES OF BASE

  $\left[H^+(aq)\right]=$$\left[OH^-(aq)\right]$
• strong acid-strong base - the initial PH is around 1, and the final PH is around 13. The equivalence point has PH roughly = to 7

  
  

  when the acid is in excess the graph has the PH of a strong acid. When the base is in excess the graph has the PH of a strong base.

  
• strong acid - weak base - the initial PH is around 1, and the final PH is around 10. The equivalence point has a PH value $<7$

  
• weak acid - weak base - the initial PH is around 3, and the final PH is around 10. The equivalence point has a PH value roughly = to 7. There is no vertical section in the PH curve, which means that there is no rapid rise in PH at the equivalence point, which thus means no acid-base indicator could work with this combination.

  
• weak acid - strong base - the PH initially rises quickly, then it flattens. the initial PH is around 3, and the final PH is around 13. The equivalence point has a PH value >7

  
• the flattened region at the start of a weak acid- strong base graph is known as the buffer region (weak acid + conjugate base)

  Half the volume required to reach the equivalence point is referred to as the half equivalence point, and at this point $\left[HA(aq)\right]=$$\left[A^-(aq)\right]$

  thus PH at this point can determine the Ka value for a weak acid:

  $Ka\ =$$\ \left[H^+\right]$

  $Ka\ =$$\ 10^{-PH}$ *
• weak acid strong base extra graph:

  
• indicators - weak acids that change colour upon dissociation. They achieve this as $Hin\ (aq)$ is a different colour to $In^-(aq)$.

  $Hin\ (aq)\ \Leftrightarrow\ In^-(aq)+$$H^+$$(aq)$

  adding acid or base will shift the equilibrium position, eventually changing the colour.
• the addition of an alkali essentially decreases the $\left[H^+(aq)\right]$. This causes the equilibrium to shift to the right in order to increase the $\left[H^+(aq)\right]$. This leads to an increase in $\left[In^-(aq)\right]$, causing the colour of the indicator to be the colour of $In^-$ (alkali colour)
• the addition of an acid essentially increases the $\left[H^+(aq)\right]$. This causes the equilibrium to shift to the left in order to decrease the $\left[H^+(aq)\right]$. This leads to an increase in $\left[HIn\ (aq)\right]$ , causing the colour of the indicator to be the colour of $HIn\ (aq)$(acid colour).
• choosing an appropriate indicator. The colour change for an indicator will occur at the Pkin/Pka value - the PH when $\left[HIn\ (aq)\right]=$$\left[In^-(aq)\right]$. An indicator which changes colour at the equivalence point/ the rapid rise in Ph/vertical section for a titration should be chosen.
• methyl orange changes colour between a PH of 3.20-4.4 (Pkin) so is a good indicator for a strong-acid weak-base titration.
• bromothymol changes colour between a PH of 6-7 (Pkin) so is a good indicator for a strong-acid strong-base titration.
• phenolphthalein changes colour between a PH of 8.2-10(Pkin) so is a good indicator for a weak-acid strong-base titration.
• buffer solution- solution that doesn't significantly change its PH when small amounts of acid or alkali are added.
• acidic buffers - made from a weak acid (such as ethanoic acid) and the conjugate base/salt of the weak acid (sodium ethanoate).
• basic buffers - made from a weak base (ammonia) and the conjugate acid/salt of the weak base (ammonium chloride).