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Lecture 3

MBIO 2360 Lecture 3: Lecture 3 – MBIO

4 Pages

Course Code
MBIO 2360
Mc Kenna Sean

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Lecture 3 – MBIO/CHEM 2360 Strong Acids and Bases  A strong acid (complete dissociation in water, no equilibrium) results in: o HCl --> H+ + Cl- o 0.1M --> 0.1M pH =1 o 0.01M --> 0.01M pH =2 o 0.001M --> 0.001M pH =3  A strong base (complete dissociation in water, no equilibrium) results in: o NaOH --> Na+ + OH- o 0.1M --> 0.1 M  [H+] = 10-14/10-1 pH=13 o 0.01M --> 0.01M  [H+] = 10-14/10-2 pH = 12 Weak Acids and Bases  Most biological acids and bases are weak o i.e. they undergo incomplete dissociation, they are in equilibrium o Draw compounds  A weak acid is a proton donor o Conjugate base is a protein acceptor  Draw compounds  A weak base is an proton acceptor o Conjugate acid is a proton donor  Biochemists choose to express all weak acids and bases as "weak acids" so: o Draw formula o Just flip the equilibrium positions  In general, draw formula and box o HA is the weak acid/conjugate acid o A- is the conjugate base  This is the acid dissociation constant and commonly expressed as: o pKa = -log10[Ka]  The weakest acids have the largest pKas  The strongest acid has the smallest pKas  If pKa is small - it'll give up its H  If pKa is big - it'll keep its H  E.g. the pKa of acetic acid is 4.76, the pKa of ammonia is 9.25  p means the -log of that number  pKa is the negative log of Ka  If pKa < 7.0, ionization is > H2O and therefore an ACID o If you are an acid and droped in water - you will dissolve and get rid of your H  If pKa > 7.0, ionization is < H2O and therefore a base Titration Curves  Analysis of titration curves provides a better insight into the properties of weak acids and bases: o Example one: a strong acid mixed with a strong base  Start with 20mL 0.1M HCl and titrate with 0.1M NaOH  Draw curve  At the start, the pH is 1, as you add more NaOH the pH goes up  Flattens out because there is no more HCl left, all NaOH  At the start there are 2 mmol of HCl (2x10-3 mol) (moles= MXL)  When 20 mL of NaOH is added, 2 mmol of NaOH is present: a. NaOH reacts with HCl b. And no acid or base is present, therefore pH is 7.0 c. One mol of acid = mol of base you have neutrality  When only 10mL of NaOH (1 mmol) is added: a. Therefore, [H+] = 1mmol/30mL = 0.033M or pH = 1.48 o Example 2 - weak acid mixed with NaOH (strong base)  At the start: 20mL 0.1M (2 mmol) CH3CCOOH and 0.1M NaOH is added  Draw curve  pH raises  Plateaus at the buffer zone  pH raises again  Plateau at the end again  At A: 20mL 0.1 M CH3COOH (2mmol) and 10 mL 0.1M NaOH (1mmol) reacts as follows:  Enter formula  At the inflection point A, there are equal amounts of the weak acid and conjugate base  This is a buffer - a mixture of a weak acid and its conjugate base a. Will be present about half way of total titrated  At B: 20mL 0.1 M CH3COOH (2mmol) and 20mL 0.1 M NaOH (2mmol)  Formula  Therefore, pH > 7.0 because only conjugate base is present  The flat zone around 10mL NaOH is the buffer zone and is most effective 0.5-1 pH units on either side of A  We will revisit the pH at A  The added OH- reacted with H+ to yield H2O a. The law of mass action then causes the HA to ionize to release more protons until the added OH- is neutralized b. Essentially Le Chatelius principle  Formula  So [HA] has decreased, [A-] increased as H+ are released from HA to neutralize incoming OH- ions. The pH eventually increases as HA becomes less abundant  The added OH- will neutralize an equal number of moles of HA  Point A a. When 10 mL of NaOH was added, the [OH-] added = 1/2 [HA] b. The acid is half neutralized and so half dissociated [HA] = [A-]  i.e. [weak acid] = [con
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