Unit 4 - INTEC Chemistry Blog | Without chemistry … · Web viewUnit 4 Author The Kong Last...
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Unit 4.4 Acid-base equilibria Define pH pH = -log 10 [H + ] no units Define K a with units For acid HA [H + ][A - ] Acid dissociation constant K a = [HA] moldm -3 Temperature dependent, K a varies with different bases Define K w with units Ionic product of water K w = [H 3 O + ] [OH − ] mol 2 dm -6 Temperature dependent, Define pK a pK a = −log 10 K a Define pK w pK w = −log 10 K w Understand the terms ‘strong’ and ‘weak’ as applied to acids and bases Difference between acid and acidic solution: • HCl acts as an acid when it dissolves in water and donates its proton to the water molecule • Solution is acidic because concentration of H 3 O + ions is greater than conc of OH - ions Strong/weak acid – fully/partially ionised Strong/weak base – fully/partially dissociate Conc No of moles of acid dissolved in a given vol of water The greater the value of K a the stronger the acid HA Strength of acid, measured by equil constant (which determines the POE) HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Equil lies so far to RHS that acid is considered to be completely ionised(strong acid) Recall the Brønsted-Lowry theory and use it to identify acids-bases Identify acid-base conjugate pairs and relate them by means of suitable equations Students are not expected to recall the Lewis theory of acids and bases Monoprotic/monobasic acid Acids that have a single proton to donate/accept Diprotic/dibasic acid Acids that have a two protons to donate/accept Acid Proton donor Base Proton acceptor Acid 1 Base 2 Acid 2 Base 1 H 2 SO 4 + H 2 O H 3 O + + HSO 4 - • Ionisation of H 2 SO 4 in water occurs in 2 stages • HSO 4 - is conjugate base of H 2 SO 4 but conjugate acid of SO 4 2- Acid 3 Base 2 Acid 2 Base 3 HSO 4 - + H 2 O H 3 O + + SO 4 2- Calculate the pH of solutions of strong acids and strong bases and of weak acids given K a Explain the action of a buffer solution and calculate its pH from suitable data based on the equations for the equilibria involved Buffer solution - Solution that maintains almost constant pH for small addition of acid or alkali A buffer solution has to contain things which will remove any hydrogen ions or hydroxide ions that you might add to it - otherwise the pH will change. An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt, eg a mixture of ethanoic acid and sodium ethanoate in solution. K a [acid] [A - ] [H + ] = [salt] pH = pK a + log [HA]
Transcript of Unit 4 - INTEC Chemistry Blog | Without chemistry … · Web viewUnit 4 Author The Kong Last...
Unit 4.4 Acid-base equilibriaDefine pH pH = -log10[H+] no unitsDefine Ka with units For acid HA [H + ][A - ]
Acid dissociation constant Ka = [HA] moldm-3
Temperature dependent, Ka varies with different basesDefine Kw with units Ionic product of water Kw = [H3O+] [OH−] mol2dm-6
Temperature dependent,Define pKa pKa = −log10 Ka
Define pKw pKw = −log10 KwUnderstand the terms ‘strong’ and ‘weak’ as applied to acids and basesDifference between acid and acidic solution:• HCl acts as an acid when it dissolves in water and donates its proton to the water molecule• Solution is acidic because concentration of H3O+ ions is greater than conc of OH- ionsStrong/weak acid – fully/partially ionised Strong/weak base – fully/partially dissociateConc No of moles of acid dissolved in a given vol of waterThe greater the value of Ka the stronger the acid HAStrength of acid, measured by equil constant (which determines the POE)HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq) Equil lies so far to RHS that acid is considered to be completely ionised(strong acid)
Recall the Brønsted-Lowry theory and use it to identify acids-basesIdentify acid-base conjugate pairs and relate them by means of suitable equationsStudents are not expected to recall the Lewis theory of acids and bases
Monoprotic/monobasic acid Acids that have a single proton to donate/acceptDiprotic/dibasic acid Acids that have a two protons to donate/acceptAcid Proton donor Base Proton acceptorAcid 1 Base 2 Acid 2 Base 1H2SO4 + H2O H3O+ + HSO4
-
• Ionisation of H2SO4 in water occurs in 2 stages• HSO4
- is conjugate base of H2SO4 but conjugate acid of SO42-
Acid 3 Base 2 Acid 2 Base 3HSO4
- + H2O H3O+ + SO42-
Calculate the pH of solutions of strong acids and strong bases and of weak acids given Ka Explain the action of a buffer solution and calculate its pH from suitable data based on the equations for the equilibria involvedBuffer solution - Solution that maintains almost constant pH for small addition of acid or alkaliA buffer solution has to contain things which will remove any hydrogen ions or hydroxide ions that you might add to it - otherwise the pH will change. An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt, eg a mixture of ethanoic acid and sodium ethanoate in solution. Ka[acid] [A - ] [H+] = [salt] pH = pKa + log [HA]Why is a mixture of nitrous acid and sodium nitrite act as a buffer solution whereas sodium nitrite solution on its own doesn’tIn a buffer both [acid] & [salt] must be large compared to the added H+ or OH- ions
Acidic Neutral AlkalinepH x (if pH is a whole number) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14[H3O+]/moldm-3 10-x moldm-3
(hydrogen ion conc)1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
NaOH(aq) Na+(aq) + OH-(aq) Kw = 10-14mol2dm-6 [OH-] =0.3moldm-3
[H3O+] x 0.3moldm-3 = 10-14 mol 2 dm-6 [H3O+] =3.33x10-14moldm-3 pH = -log10(3.33x10-14) = 13.5HCl(aq) pH of 3, hydrogen ion conc? Molarity of acid? pH =3 [H3O+] = 10-3 = 1 x10moldm-3 Assuming the acid to be 100% ionised, acid being monoprotic, molarity of acid also 1 x10-3moldm-3
HNO2 is a weak acid Ka = 4.7 x10-4moldm-3 HNO2(aq) + H2O(l) H3O+(aq) + NO2-(aq)
Ka? [ H 3O + ]x[ NO 2
- ] Ka = [HNO2]
Calculate pH of 0.12moldm-3 nitrous acid solution[H+] = [NO2
-] so [H+]2 = Ka x[HNO2] √Ka x0.12 = 0.00751moldm-3 pH =-log10[H+] =2.12Calculate pH of buffer solution made by adding 1.38g of sodium nitrite NaNO2 to 100cm3 of 0.12moldm-3 nitrous acid solution,
Ka = 4.7 x10-4moldm-3
Moles NaNO2 =1.38/69 =0.02, [NO2-] =0.02/0.1 =0.2moldm-3
Ka[acid] 4.7 x 10 -4 x0.12 [H+] = [salt] = 0.2 =2.82x10-4, pH =-log10 2.82x10-4 =3.55A buffer solution contains HA(aq) at a concentration of 0.300 mol dm–3, and its sodium salt, NaA, at a concentration of 0.600 mol dm–3. Calculate the pH of this buffer solution.
pH = pKa + log[HA]
][A –= log10 (5.62 105) + log10 ]300.0[
]600.0[ = 4.55
Understand the principles involved in acid-base titrationsRecall the sketch curves for the variation in pH during the following titrationsstrong acid – strong base weak acid – strong basestrong acid – weak base
Strong acid and strong base, running acid into alkali Strong acid and strong base, running alkali into acidLarge change in pH for a very small addition of acid and around the end point of the titration
Strong acid and weak base, running acid into alkali Weak acid and strong base, running acid into alkaliAt equivalence point all acid and base react to make a salt
An aq solution of ethanoic acid of conc 1.0moldm-3 has a pH of 2.8 Sketch how the pH changes during the titration of 25cm3 of 1moldm-3 aq ethanoic acid with aq NaOH of the same conc
25cm3 of 1moldm-3 HA(aq)(weak acid) was titrated with 1moldm-
3 NaOH(aq) and the pH measured throughout. Use titration curve to find pH at end point of titration 9.0-9.4Find range of pH values over which the mixture acts as a buffer 5.2-5.8
Use titration curves to determine Ka for a weak acid
Explain the choice of a suitable indicator for an acid-base titration given pKInd Students will not be expected to recall the pH ranges of indicators other than methyl orange and phenolphthaleinHIn + H2O H3O+ + In- [In - ][ H 3O + ]
KIn = [HIn]
Indicator pH range(working range) Acid colour Alkaline colourMethyl orange 3.2-4.5 Red Yellow Thus suited for strong acid/weak base titrationsPhenolphthalein 8.2-10.0 Colourless Magenta Thus suited for strong base/weak acid titrationsBromothymol blue 6.0-7.0 Yellow Blue
Demonstrate an understanding of how the value of enthalpy of neutralisation is related to the strength of acids and basesEnthalpy of neutralisation HN Enthalpy change when one mole of water is formed from reaction of an acid with a baseEnthalpy of neutralisation is H for: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
½H2SO4(aq) + NaOH(aq) ½Na2SO4(aq) + H2O(l)CH3CO2H(aq) + NaOH(aq) CH3CO2Na(aq) + H2O(l)
HN quite constant at -56kJmol-1 when acid and base are both strongIf acid and base are fully ionised, ionic equation for all these reactions is the same: H+(aq) + OH-(aq) H2O(l)If either the acid or base is weak, however, some of the energy is absorbed in order to ionise the weak acid or base(amount of heat liberated less than -57kJmol-1) HN of some acids and bases are sometimes more exothermic than -57kJmol-1 because there is also heat liberated on hydration of ions once formed. Energy released on hydration of ions is sometimes greater than energy required to complete ionisation of the acid or base
