Buffers

BuffersPrepared byJulius Victorius A. SaluriaBuffersSolutions composed of a weak acid/base and its conjugate salt.i.e. HOAc and NaOAc in one solution or NH3 and NH4Cl in one solution.Solutions that can resist drastic changes in pH when a strong acid or a base is added to it.pH is calculated using the Henderson-Hasselbalch EquationHenderson-Hasselbalch EquationWhere pKa is log of Ka or the acid dissociation constant of the weak acid HA, b is the number of moles of the basic component (the conjugate base of the weak acid) and a is the number of moles of HA, both in equilibrium.

Henderson-Hasselbalch EquationWhere pKb is log of Ka or the base dissociation constant of the weak base B, b is the number of moles of B and a is the number of moles of the acid component (the conjugate acid) HB+, both in equilibrium.

Buffer PreparationHA/NaA or the B/BH+Cl- bufferPrepared by combining a weak acid/ weak base with the salt of its conjugate base/acid.Partial NeutralizationPrepared by generating the conjugate species from the neutral species by adding a strong base to HA or a strong acid to B. The added acid/base should have a lower number of moles compared to that of the weak species.Buffer Preparation: The HA-NaA BufferFor example:1.00 L of 1.00 M buffer with pH 4.500 is to be prepared from 2.00 M HOAc (Ka = 1.85x10-5) and NaOAc solids (82.00 g/mol). How much of each component should be mixed to prepare the desired buffer?Step 1Obtain the total number of moles of the buffer (moles of the buffer is equal to the sum of the number of moles of a and b)Moles buffer = (Vbuffer)(Mbuffer)Moles buffer = (1.00 L)(1.00 M)Moles buffer = 1.00 molea + b = 1.00 mole (Eqn 1)Buffer Preparation: The HA-NaA BufferStep 2Determine the ratio of the base component to the acid component using the Henderson-Hasselbalch EquationpH = pKa + log (b/a)log (b/a) = pH pKa = 4.500 4.733 = -0.233(b/a) = 10^(-0.233) = 0.585Or b = 0.585a (Eqn 2)Buffer Preparation: The HA-NaA BufferStep 3Combine Equations 1 and 2 to obtain values of a and ba + b = 1.00; b = 0.585aTherefore: a = 0.631 moles and b = 0.369 molesBuffer Preparation: The HA-NaA BufferStep 4Determine the amount of the acid and base components needed. If the reagent is liquid, it should be expressed either in L or mL; if solid express in grams or mg.V HOAc = a/MHOAcV HOAc = 0.3155 L or 316 mLmass NaOAc = b*FMHOAcmass NaOAc = 30.3 gramsBuffer Preparation: The HA-NaA BufferTherefore, to prepare the HOAc-NaOAc buffer that is 1.00 L, 1.00 M and is at pH 4.500, one should mix 316 mL of 2.00 M HOAc with 30.3 grams of NaOAc solids then dilute the solution to 1.00 L.Buffer Preparation: The HA-NaA BufferBuffer Preparation: HA-NaA BufferDRILL!Suppose you are to prepare a 0.250M, 50.0 mL buffer with pH 9.00 from 1.0 M NH3 (Kb = 1.85x10-5) and NH4Cl (53.45 g/mol) salt.How much of each component do you need to prepare this buffer?Buffer Preparation: Partial NeutralizationFor example:1.00 L of 1.00 M buffer with pH 4.500 is to be prepared from 2.00 M HOAc (Ka = 1.85x10-5) and 1.00 M NaOH. How much of each component should be mixed to prepare the desired buffer?

Step 1Obtain the total number of moles of the buffer (moles of the buffer is equal to the sum of the number of moles of a and b and for buffers prepared by partial neutralization, a + b is equal to the number of moles of the weak species)Moles buffer = (Vbuffer)(Mbuffer)Moles buffer = (1.00 L)(1.00 M)Moles buffer = 1.00 molea + b = 1.00 mole (Eqn 1)moles HOAc = 1.00 moleBuffer Preparation: Partial NeutralizationStep 2Determine the ratio of the base component to the acid component using the Henderson-Hasselbalch EquationpH = pKa + log (b/a)log (b/a) = pH pKa = 4.500 4.733 = -0.233(b/a) = 10^(-0.233) = 0.585Or b = 0.585a (Eqn 2)Buffer Preparation: Partial NeutralizationStep 3Combine Equations 1 and 2 to obtain values of a and ba + b = 1.00; b = 0.585aTherefore: a = 0.631 moles and b = 0.369 molesNOTE: The important quantity here is b (or the number of moles of OAc-) because this is equal to the number of moles of NaOH needed to partially neutralize HOAc.Buffer Preparation: Partial NeutralizationWhy is it so?

The neutralization reaction is:HOAc + OH- OAc- + H2OAnd because it is partial neutralization, OH- should be the limiting reactant.Using the ICE Table, hence moles OH- is 0.369.

Buffer Preparation: Partial NeutralizationHOACOH-OAc-Initial1x0Change-x-x+xEquilibrium0.63100.369Step 4Determine the amount of the acid and base components needed. If the reagent is liquid, it should be expressed either in L or mL; if solid express in grams or mg.V HOAc = a/MHOAcV HOAc = 0.3155 L or 316 mLV NaOH= b/MNaOHV NaOH = 0.369 L or 369 mLBuffer Preparation: Partial NeutralizationTherefore, to prepare the HOAc-NaOAc buffer that is 1.00 L, 1.00 M and is at pH 4.500, one should mix 316 mL of 2.00 M HOAc with 369 mL of 1.00 M NaOH then dilute the solution to 1.00 L.Buffer Preparation: Partial NeutralizationDRILL!Suppose you are to prepare a 0.250M, 50.0 mL buffer with pH 9.00 from 1.0 M NH3 (Kb = 1.85x10-5) and1.0 M HCl.How much of each reagent do you need to prepare this buffer?Buffer Preparation: Partial NeutralizationBuffer CapacityThe amount (number of moles) of the strong acid or the strong base that the buffer can neutralize.Remember that the buffer is composed of a weak species and its conjugate salt.If a strong acid is added, the base component of the buffer reacts with the acid; on the other hand, if a strong base is added, the acid component of the buffer reacts with the base.The Revised Henderson-Hasselbalch Equation: Calculating pH of Buffer Solution After the Addition of H+ or OH-Where s is the amount (in moles) of the strong base or strong acid that is added.The term b+s is paired with a-s. This is the case when a strong base is added.The term b-s is paired with a+s. This is the case when a strong acid is added.What is the form of the Revised HH Equation in terms of pOH?

DRILL!What is the pH of a 50.0 mL, 0.25M, pH 6.00 buffer (pKa of the acid component is 5.25) when 1.00 mL of 0.100 M HCl is added?What is the pH of a 50.0 mL, 0.25M, pH 6.00 buffer (pKa of the acid component is 5.25) when 1.00 mL of 0.100 M NaOH is added?The Revised Henderson-Hasselbalch Equation: Calculating pH of Buffer Solution After the Addition of H+ or OH-