A strong acid is completely protolysed

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Elmer Lambert
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1 Strong acid A strong acid is completely protolysed Ex mol HCl to 1 liter H 2 O (=0.16 M) [H 2 O] = 55,6 M 1. Set up the reaction. HCl is a strong acid. 2. Amount acid resp. base before and after equilibration HCl + H 2 O Cl - + H 3 O + Before equil. 6 55,6 noll 10-7 After equil. noll 55, ph = -lg 6 ph = 0,8 Compare with ph in stomach (1-3). Strong acid Ex. Calculate ph in a solution of HCl with a conc. of M. HCl is a strong acid that gets proteolysed completely and (HCl) = (H 3 O + ) = M. 1. Set up the reaction. 2. Write amount acid and base before and after equilibration. HCl + H 2 O Cl - + H 3 O + Before equil. 0,003 55,6 noll 10-7 After equil. jvt noll 55, ph = -lg H 3 O + = ph = -lg ph = 2.5 1

2 Before equil. noll 10-7 After equil. - X X X X 2:a grade equation... Make two approximations: 1) X << 2) X >> 10-7 Before equil. noll 10-7 After equil. - X X X X 2

3 Before equil. noll 10-7 After equil. - X X X X X K a HAc =1, X Before equil. noll 10-7 After equil. - X X X X X X K a HAc =1, , = X2 3

4 Before equil. noll 10-7 After equil. - X X X X X X K a HAc =1, , = X2 X = 1, M = [H 3 O + ] ph = 2,88 Ion product of water Ex. Calculate ph in a solution of 0.17 M NaOH? NaOH is a strong base which is completely proteolysed. Every mole NaOH will release OH ions => [OH - ] = 0.17 M. NaOH + H 2 O H 2 O + Na + + OH - Calculation of [H 3 O + ] and [OH - ] and thereafter ph with: Kw Kw=[H 3 O + ] [OH - ] [H 3 O + ]= [OH - ] and Kw= is added in the equation gives => [H 3 O + ]=10-14 /0.17 [H 3 O + ]=5.88 x ph=-lg[h 3 O + ]=13.23 Answer:

5 Ion product of water Ex. Calculate ph I 0.17 M NaOH? Alternative way. [OH - ] = 0.17 M. Calculate ph from poh. pkw = ph + poh poh= -lg[oh - ] = -lg[0.17] = 0.77 ph+poh=14 ph = 14 poh poh=0.77 => ph = =13.23 Answer Buffer Buffer equation How does ph change in a buffer upon dilution Exempel: 0,2 M acetatbuffert ( mol HAc, mol Ac -, 1 liter H 2 O) ph = pk a + lg ph = pk a + lg 1 ph = 4,76 Tillför 1 liter vatten 0,05 ph = pk a + lg 0,05 0,05 0,05 ph = pk a + lg 1 ph = 4,76 5

6 3. Buffer Imorrn Buffer equation What is ph when the buffer is buffering against acid: Ex. What is the ph in a 0.2 M acetatbuffer with addition of acid? (pka HAc = 4,76) (at pka there is 0.1 mol HAc, and 0.1 mol Ac -, in 1 liter H 2 O) ph = pk a + lg Add 0,05 mol saltsyra (50 mmol H 3 O + ) [Ac - ] H 3 O + + Ac - H 2 O + HAc Before 0,05 After 0 0,05 5 ph = 4,76 + lg 0,05 5 ph = 4,76 + lg 0,33 ph = 4,28 3. Buffer Buffer equation Vid pka är HAC/Ac - 50:50 x ekvivalenspunkt When the buffer is buffering, the equilibrium is changed: After addition HCl x HAC/Ac - 50:50 ph=4,28 6

7 3. Buffer Buffer equation When the buffer is buffering against a base: Exempel: What is the ph if base is added to 0.2 M acetate buffer? pka (HAc) = 4,76 ph = pk a + lg [Ac- ] ( mol HAc, mol Ac -, 1 liter H 2 O) Add 0,05 mol NaOH (50 mmol OH - ) OH - + HAc H 2 O + Ac - before 0,05 after 0 0,05 5 ph = 4,76 + lg 5 0,05 ph = 4,76 + 0,4771 ph = 4,76 + lg 3 ph = 5,24 3. Buffer Buffer equation When the buffer is buffering, the equilibrium is shifted: HAc + OH - Efter tillsats HCl Efter tillsats OH - x x Ac - + H 2 O Ac - x ekvivalenspunkt 7

8 4. Titration Ex. 20 ml HCl is titrated with 16 ml of 0.1 M NaOH to the equivalenspoint. What is the concentration of HCl? An acid with one proteolytic step is titrated against a strong base. At the equivalenspoint, the n(acid) in the solution is the same as the amount (n) of base added. How much base do we have? We use n=cv => 0.1M x L = 1.6 x 10-3 mol = mol At the equivalenspoint the n(acid) = n(base) => Which means that n(acid)= mol. What is the concentration of acid? c = n/v = mol / 0.02 L = 0.08 M. equivalenspoint Answer: The concentration of the HCl solution is 80 mm. 5. Physiologic ph regulation Bicarbonate buffer Bicarbonate buffer, closed system. What is the ph after added acid? 1. Set up reaction What is conj. acid and base? Write before and after CO 2 + 2H 2 O H 2 CO 3 +H 2 O HCO 3- + H 3 O + 2. What formula should we use? Bicarbonate buffer = Buffer formula [HCO 3- ] ph = pk a + lg [CO 2 ] + [H 2 CO 3 ] H 3 O + + HCO 3-2H 2 O + [syra] Before ,2 After ,2 ph = 6,1 + lg 19 6,2 ph = 6,1 + 0,49 ph = 6,59 8

9 5. Physiologic ph regulation Bicarbonate buffer Bicarbonate buffer in an, open system, what is ph after addition of acid? CO 2 + 2H 2 O H 2 CO 3 +H 2 O HCO 3- + H 3 O + acid = 0,23 5,3 = 1,2 mm 24 mm syra 5 mmol/l ph = pk a + lg [HCO 3 - ] [CO 2 ] + [H 2 CO 3 ] ph = 6,1 + lg 19 1,2 H 3 O + + HCO 3- H 2 O+ H 2 CO 3 ph = 6,1 + 1,2 Before ,2 After ,2 Answer: ph = 7,3 Breath 1,2 9

2. Acid base concept -calculations. 3. Buffert -calculations. 5. Physiologic ph regulation -calculations

Acid Base lecture 15-16 September 2014 Agnes Rinaldo-Matthis 1. Introduction to ph 2. Acid base concept -calculations 3. Buffert -calculations 4. Titration 5. Physiologic ph regulation -calculations 6.