Aya Alomoush. Amani Nofal. Mamoon AhraM

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Marsha Eaton
5 years ago
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1 4 Aya Alomoush Amani Nofal Mamoon AhraM

2 Hi doctors, in this sheet we will continue talking about ph,buffer, titration curve, soooo lets begin our party. 1)Handerson-Hasselbalch equation (Important ) *PH =Pka when : PH =Pka + log([a-]/[ha]) [A-]=[HA] (remember log(1) =0) or in another expression when 50% of the acid is dissociated into its conjugate base.(this means 50% remains as acid and 50% become conjugate base) *PH>Pka when : ( log([a-]/[ha]))>0 or in another way when the concentration of the conjugate base is higher than that of its acid (]A-[>]HA[) (more than half of the acid is dissociated into its conjugate base ). *PH<Pka when: (log([a-]/[ha]))<0 or in another way when the concentration of the acid is higher than that of its conjugate base(]ha[>]a-[) ( when less than the half of the acid is dissociated into its conjugate base). REMEMBER :: -WHEN PH< PKa predominates H+ on( substance protonated) the protonated (acid )form -When ph>pka H+ OFF (substance deprotonated), the deprotonated (conjugate base) form predominates. 2) Le Châtelier s principle: (When any system at equilibrium is subjected to a change then the system readjusts itself to counteract the effect of the applied change and a new equilibrium is established.) 1 P a g e

3 *example (from the slides ): -In the other words : if we add more reactants the equilibrium shifts to right, if we add )نفس التوجيهي ) left more products the equilibrium shifts to And Le Châtelier s principle is important to explain the mechanism of buffering (which is the next topic ). )هو المحلول المنظم تبع التوجيهي 3)Buffers) Buffers are solutions that resist changes in ph by changing reaction equilibrium when small to moderate amounts of a strong acid or base is added. They are composed of a weak acid and its salt (conjugate base),also they can be composed of a weak base and its salt (conjugate acid ). -Examples of buffers : Acid/its salt(conjugate base) a. CH3COOH /CH3COONa b. H3PO4/NaH2PO4 c. H2PO4-/K2HPO4 d. H2CO3/KHCO3 e. NH3/NH4CL 2 P a g e

4 *How do buffers work? As we all already know they work by changing the reaction equilibrium. Let's talk about what actually happens when we add a strong acid or base to a buffer. a. Addition of a strong acid : If a strong acid is added to a buffer,the H+ from the strong acid will react with the conjugate base of the weak acid formig more of the acid( H+ + A- HA). Since the added H+ is consumed by this reaction, the ph will decrease only slightly (according to Handerson-Hasselbalch equation ). b. Addition of a strong base : If a strong base is added to a buffer, the weak acid will give up its H + in order to transform the base (OH - ) into water (H2O) and the conjugate base: HA + OH - A - + H2O. Since the added OH - is consumed by this reaction, the ph will increase only slightly (according to Handerson-Hasselbalch equation ). *Titration curve of a buffer ( CH3COOH/CH3COO- (an example from the slides )) Equivalence point Addition of a strong acid Mid point * x Buffering capacity Addition of a strong base 3 P a g e

5 a. What is the midpoint? it is the point at which the PH=Pka ((log([a-]/[ha])=0 because [A-]=[HA])and in this example it is equal to 4.76 (for this acid). b. The buffer stops resisting the change(increase or decrease ) in PH when there is no more conjugate base to react with H+ in the case of addition of a strong acid or when there is no more acid to react with OH- in the case of addition of a strong base. c. What is the buffering capacity? it is the region of PH at which the Buffer resist changing in the PH (all buffers have the same range of buffering capacity which is( Pka+- 1),but ( you should remember that different buffers have different Pka)and in this example the buffering capacity range is (4.76-1=3.76,4.76+1=5.76) ALSO the buffering capacity depends on Buffer concentration (positive relationship) The Pka of the buffer The desired PH d. What is the equivalence(end) point? the point at which the acid is exactly neutralized (when the equivalents of the acid equal the equivalents of the base (remember that the equivalence=normality *volume)(normality*n*=n*m) in this example the equivalence of both the acid and the base at this point = 1 e. What is the ratio of the conjugate base to the acid ([CH3COO-]/[CH3COOH]) at :(CONSIDER THAT Pka=4.8): PH =3.8? the ratio would be 1/10 according to Handerson-Hasselbalch equation. - PH PKa = -1 =log (]A-[/]HA[) (]A-[/]HA[) = 1/10 PH=5.8?? the ratio would be 10/1 according to Handerson-Hasselbalch equation. - PH PKa = +1=log (]A-[/]HA[) (]A-[/]HA[) = 10/1 Mid point : the ratio will be 1/1 Point *(approximately): it is a point in the middle between the midpoint and PH=3.8 so the be in between the two ratios which will be 1/5. 4 P a g e

Point X (approximately): it is a point in the middle between the midpoint and PH=5.8 so the answer will be in between the two ratios which will be 5/1. *How do we choose buffers?

6 Point X (approximately): it is a point in the middle between the midpoint and PH=5.8 so the answer will be in between the two ratios which will be 5/1. *How do we choose buffers? We choose buffers depending on the desired PH (the desired PH should be in the range of the buffering capacity ). *What is the most suitable buffer from below if the desired PH PH=5? acetate buffer. PH=7? phosphate buffer. PH=9? Ammonium buffer. * Questions from the lecture : a. A solution of 0.1 M acetic acid and 0.2 M acetate ion. The pka of acetic acid is 4.8. Hence, the ph of the solution is given by : PH = 4.8+LOG(2)=5.1 (directly by Handerson- Hasselbalch equation) b. Predict then calculate the ph of a buffer containing 0.1M HF and 0.12M NaF? (Ka = 3.5 x 10^-4 ) Pka=-log(3.5*10^-4)=3.4 PH=3.4+LOG(0.12/0.1)= P a g e

7 C. 0.1M HF and 0.1M NaF, when 0.02M HCl is added to the solution? [NaF]= =0.08, [HF]= =0.12 PH=Pka+log(0.08/0.12)=3.22 d.what is the ph of a lactate buffer that contain 75% lactic acid and 25% lactate? (pka = 3.86) *PH=3.86+LOG(25/75)=3.38 e. What is the concentration of 5 ml of acetic acid knowing that 44.5 ml of 0.1 N of NaOH are needed to reach the end of the titration of acetic acid? Also, calculate the normality of acetic acid? Ammmm the end of the titration =the equivalence point - N*VOLUME OF acetic acid =N*VOLUME of NaOH N*5*10^-3=44.5*0.1 *10^-3 THEN N=0.89 =MOLARITY (since acetic acid is monoprotic acid ) -THIS LINK EXPLAINS THE WAY THESES EXCERSISES ARE ANSWRERED axdpe X98r-E 6 P a g e

8 *Titration curve of a polyprotic acid (H3PO4 as an example ) NOTES ON THIS CURVE : There are 3 buffering capacities with three different mid points. The strength of the acid decreases as FOLLOW (H3PO4>H2PO4->HPO4-2). Value 0.5 at the x axis is the first midpoint,the value 1.0 on the same axis is the first equivalent point while the value 2 on the same axis is the second equivalence point. Remember that at any equivalence (end )point all the acid is neutralized by the base so we don't have any base or acid. - the best video: 7 P a g e

9 Summary: In aqueous solution, the relative concentrations of a weak acid and its conjugate base can be related to the titration curve of the acid. In the region of the titration curve in which the ph changes very little upon addition of acid or base, the acid/base ratio varies within a narrow range (10:1 at one extreme and 1:10 at the other extreme). Buffer solutions are characterized by their tendency to resist ph change when small amounts of strong acid or strong base are added. Buffers work because the concentration of the weak acid and base is kept in the narrow window of the acid titration curve. Many experiments must have a buffered system to keep a stable ph. Past Papers 1- If the ph of a solution decreased from 7.5 to 7. What change has occurred to the concentration of H3O+? a- increased 3 times b- Increased 5 times c- Increased 500 times d- Increased 10^5 times e- Increased 10^ (1/2) times 2- If you have X moles of KOH, how many moles of an acid must be added to have a buffer with equal concentrations of A- and HA? a- X b- X/2 c- 2X d- 1.5 X e- None of the above 8 P a g e

10 3- Given pka of different acids, which one will have the strongest conjugate base when being dissociated with water? a- 3.5 b- 2.9 c d- 7.2 e ml of a buffer has a concentration of 0.2 M. The buffer is composed of a weak acid component and a conjugate base component and its ph=7.57. If 1 ml of 1 M HCl is added, what will be the new ph value? (Pka=7.57) a-7.5 b-8 c-7 d-not changed 5- Below is the pka of some weak acids. Which weak acid will be 91 % undissociated at ph=4.86? a- Acetoacetic acid pka = 3.6 b- Lactic acid pka=3.9 C- beta-hydroxyl butyric acid pka=4.8 d- propionic acid pka=4.9 e- Imidazolium pka= A phosphate buffer is composed of 0.5 M Na2HPO4 and 0.25 M NaH2PO4. If 0.05 M of HCL are added, what would be the approximate ph if pka=7.2. a-7.3 b-6.5 c-6.9 d A patient was found to have undetected diabetes mellitus for a while, in the urine sample taken [HCO3-] = 14.1 and [CO2] = 1.1, Pka for blood=6.1, most likely ph of blood was: A) 7.1 B) 7.2 C) 7.4 D) 7.5 E) P a g e

11 8. The pka of a base is 4. If you have a 0.01M solution of this base, what is the ph? A) 8 B) 9 C) 10 D) 11 E) Gastric juice (ph= 1.4) compared to human s blood (ph= 7.4): A) [H+] in gastric juice is 6 times higher than in blood B) [H+] in gastric juice is 10^6 times higher than in blood C) [H+] in blood is 10^6 times higher than in gastric juice D) [H+] in gastric juice is 7 times higher than in blood Answers A C E A E A B D B Q1+Q9- Remember this rule: ( The difference in [H+} = 10^(pHx - phy). The end)sorry for any mistake) Stay Blessed ^_^ 10 P a g e

Acids and bases, ph and buffers. Dr. Mamoun Ahram Lecture 2

Acids and bases, ph and buffers Dr. Mamoun Ahram Lecture 2 ACIDS AND BASES Acids versus bases Acid: a substance that produces H+ when dissolved in water (e.g., HCl, H2SO4) Base: a substance that produces