point is a buffer (since [HA] = [A]). From the graph above it can be seen that adding OH- does not significantly affect thepH in the region from about 8 mL to 18 mL. The pH in this region is intermediate between that of the weak acid at the startof the titration and the pH at the equivalence point.In any solution containing a conjugate acid/base pair (HA and A-), this equilibrium occurs:НА(аq) +H2O(1)А (аq) +H3O*(aq)[H3O*][A¯][HA]for which the equilibrium constant is Ka(where stronger acids have larger Ka values)[A]which can be rewritten as: pH = pka + log10HAT) where pKa =-log10(Ka).[equation 1]An effective buffer solution needs to be able to consume added acid and base so the concentrations of the acid and conjugatebase in the buffer need to be approximately equal: [A¯V[HA] - 1 which means the pH of the buffer will be close to the pKaof the acid used to prepare it (log10(1) = 0 in equation 1).When a small quantity of a strong acid is added to a buffer, it reacts with the A. For example, say you have a buffer madefrom a weak acid with pKa = 5.20 and the concentrations of the weak acid and base are: [HA] = 0.130 M and [A¯] = 0.130M. The pH of this buffer will be 5.20 (from equation 1). If you add 1.0 mL of 0.10 M HCI (which contains 0.00010 mol H*)to 100 mL of the buffer the following reaction happens:A(aq) +H30*(aq, added) →HA(aq) + H2O(1)Initial quantities0.01300.000100.0130molFinal quantities0.01290.0131molHence, the amount of HA increases slightly while that of A is decreased by the same amount. The H* from the strong acidis gone. The total volume of the solution is 101.0 mL and the concentrations are now: [HA] = 0.0131 mol/0.101 L= 0.1297M and [A-]=0.0129 mol/0.101 L = 0.1277 M and the pH = 5.20 + log10(0.1277/0.1297) = 5.19. For comparison, if you add1.0 mL of 0.10 M HCl to 100 mL of water, the pH of the solution changes from 7.0 to 3.0 (the actual concentration of H*increases by a factor of about 10,000 times).Buffers can be prepared for almost any pH by appropriate choice of a weak acid and conjugate base pair. Since a buffer ismost efficient at a pH close to the pKa of the acid (when the weak acid and conjugate base concentrations are approximatelyequal), this provides a way to select the appropriate pair for the desired pH of the buffer. The following table gives somecommon weak acid/base conjugate pairs and the pH range over which they can be used to prepare buffers.Acid/Base Pairpka of AcidpH Range of Effective BuffersH3PO4/H2PO4-2.151.1 to 3.1HF/F-3.172.2 to 4.2НСООН/НСО3.742.7 to 4.74.76CH3COOH/CH3CO2-H2PO4-/HPO42-NH4*/NH3HCO3-/CO32-HPO42-/PO433.8 to 5.87.205.9 to 7.99.248.2 to 10.210.339.3 to 11.312.3811.4 to 13.4Buffers are most effective at pH values in the range pka -1 to pKa + 1, e.g., a buffer based on NH4*/NH3 will only maintaina stable pH in the range 8.24 to 10.24. The further the pH of the buffer is away from the pKa value, the less effective it isbecause the concentrations of weak acid and conjugate basc must be uncqual to produce such pH's. PRE-LAB CALCULATION(Compulsory, must be completed to be allowed to start experiment. To be done before your lab class.)The tables below show the solutions you will have in the laboratory and the pH of the buffer you will need to prepare (basedon your course and workspace). Select the reagents to prepare your buffer (refer to the table of pka values on the previouspage) and in the space below calculate the volumes of cach solution required to prepare 20.0 ml. of the buffer. Circle thebuffer pH and reagents you will use.Solutions available to prepare buffers0.10 M CH3COOH0.10 M NaH2PO40.10 M NH4CI0.10 M NaHCO3WorkspaceBuffer pHBuffer pH0.10 M CH3COON0.10 M NazHPO40.10 M NH30.10 M NaɔCO3(CHEM1011)(CHEM1031/CHEM1051)My selected Buffer1 and 74.764,50pH2 and 87.207.303 and 69.249.404 and 910.3310.405 and 104.764.601. Choose the acid/base pair you will be using to make your buffer in the laboratory. Refer to the table of pKa's in theintroduction and the table of solutions available in the laboratory.0.10 M NHẠCI0.10 M NH32. Calculate the volumes of cach solution you will need to mix to make your buffer.Use the pH of your buffer and the pk, of the acid component of your buffer to calculate [A¯[HA].pH = pka + logio(HA]) which gives: log1o(HA = pH - pKa =which means that(HA]Since the concentrations of the acids and conjugate bases are all the same in the solutions provided, the ratio(A-V[HA] equals the ratio of the volume of base (V½) to the volume of acid (V,) (remember, A¯ is the base).[A/HA] = V½/V½ = ,and to simplify the algebra we will call this ratio 'C.Volume of buffer = 20.0 mL., so V, + V½ = 20.0 or /, = 20.0 – Vz-VWV2 =C=(20.0– VY) so that Cx V =(20.0 – V) which gives Va= 20.0 /(1+ C). Use this to calculate thevolumes of acid and base solutions required to make 20.0 mL of your assigned buffer.Volume of acid (V)=mL.Volume of base (V) =mL

We have to find the volume of acid and volume of base, to make given buffer Given pH = 9.24 pKa =…

Answered: ume of acid (V) = ml. Volume of base ()…

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Chemistry

ume of acid (V) = ml. Volume of base () = ml.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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