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AQA A2 acids
| Question | Answer |
|---|---|
| Write the full equation between Calcium carbonate and hydrochloric acid | 2 HCl + CaCO3 CaCl2 + CO2 + H2O (remember to check the formula) |
| Write the ionic equation between the solutions of sodium carbonate and hydrochloric acid | 2H+ + CO3-2 CO2 + H2O (remember to check the formula) |
| Write the ionic equation between the solutions of potassium hydroxide and hydrochloric acid | H+ + OH- H2O |
| Write the ionic equation between the solutions of magnesium and hydrochloric acid | Mg + 2H+ Mg+2 + H2 |
| Write the ionic equation between the solutions of magnesium oxide and hydrochloric acid | MgO + 2H+ Mg+2 + H2O |
| Name the two conjuguate pairs in this reaction: HCl + H2O H3O+ + Cl-; for each pair, indicate which species is the acid, which is the base | HCl (acid) and Cl- (base); H2O (base) and H3O+ (acid) |
| Name the two conjuguate pairs in this reaction: NH3 + H2O NH4+ + OH-; for each pair, indicate which species is the acid, which is the base | H2O (acid) and OH- (base); NH3 (base) and NH4+ (acid) |
| Identify the acid-base pairs in the reaction: HIO3 + H2O H3O+ + IO3- | H2O (base) and H3O+ (acid); HIO3 (acid) and IO3- (base) |
| Ka for Propanoic acid is 1.35x10^-5 and for Methanoic acid 1.78x10^-4; complete the equation CH3CH2COOH + HCOOH | Methanoic acid has the highest Ka so will dissociate more easily so it will be the acid; whereas Propanoic acid will be the base; CH3CH2COOH + HCOOH CH3CH2COOH2+ + HCOO- |
| Kw=1.0x10^-14 at 25C; what is the pH of water at 25C? | Kw=[H+][OH-]; here [H+]=[OH-] so pH=-log[H+]=-log(square root Kw); remember to type brackets on your calculator |
| H2O H+ + OH-; delta-H=+57kJ.mol^-1 and Kw=1.0x10^-14 at 25C; would the pH of water at 50C be lower or higher than at 25C? | as temp increases, the equilibrium will shift to the right as forward is endo; so CONCENTRATION of H+ increases; Kw will increase so pH=-log[H+] will decrease |
| Why is water neutral at 37C, even though the pH is 6.87? | there are equal numbers of moles of H+ and OH- |
| How do you work out the pH of a strong acid given [HX]? | strong acid= fully dissociates so [H+]=[HX]; pH=-log[H+]; BEWARE of H2SO4 and other DI-protic acids for which [H+]=2x[H2X]!!! |
| How do you work out the pH of a strong (monoprotic) base given [base]? | strong base= fully dissociates so [OH-]=[base]; using Kw, [H+]=Kw/base; pH=-log[H+] |
| How do you work out the pH of a strong (diprotic) base given [base]? | strong base= fully dissociates BUT for each mole of the base, 2 OH- will be produced so [OH-]=2x[base]; using Kw, [H+]=Kw/base; pH=-log[H+] |
| Calculate the pH of a 0.1mol.dm-3 solution of NaOH | [OH-]=[NaOH]=0.1mol.dm^-3; [H+]=Kw/[OH-]=1x10^-14/0.1=1x10^-13; pH=-log[H+]=13 |
| What is the expression of Ka and units for Ethanoic acid? | Ka=[CH3COO-]x[H+]/[CH3COOH]; water is in excess so does not appear in Ka; units are mol.dm^-3 |
| How do you work out the pH of a weak acid given [HX] and Ka? | we assume that 1) [H+]=[X-] and 2) [HX] hardly changes so after rearranging expression of Ka, [H+]=square root(Ka x [AH]); pH=-log[H+] |
| Calculate the pH of a 0.1 mole.dm-3 Ethanoic acid with Ka=1.7x10^-5 mol.dm^-3 | 2.48 |
| Calculate the pH of a 0.65 mole.dm-3 Ethanoic acid with Ka=1.7x10^-5 mol.dm^-3 | 4.39 |
| Calculate the pH of a 0.044 mole.dm-3 HClO with Ka=3.7x10^-8 mol.dm^-3 | 5.32 |
| Calculate the pKa of a weak acid when a 0.13mol.dm^-3 has a pH of 3.52 | Ka=7.02x10^-7 and pKa=6.15 |
| Calculate the pKa of a weak acid when a 0.078mol.dm^-3 has a pH of 5.19 | Ka=5.34x10^-10 and pKa=9.27 |
| Chloroethanoic acid has a pKa of 2.88 and ethanoic acid of 4.77. Which is the strongest acid? | Chloroethanoic acid; the smaller the pKa, the stronger the acid |
| Ka of Benzoic acid is 6.3x10^-5 and of Hydrocyanic of 4.9x10^-10. Which is the strongest acid? | Benzoic acid; the larger Ka, the more dissociated, the stronger the acid |
| What is the equivalence point? | the point where volume of one solution has reacted exactly with the volume of the second solution (moles of H+=moles OH-); the point in the middle of the vertical section of the titration curve |
| What is the end point? | the point for which the indicator changes colour; the point for which there is equal amounts of the weak acid and its conjuguate base form of the indicator |
| What are the conditions to choose an indicator? | the end point must be the same as the equivalence point so the indicator must change colour within the vertical section of the pH curve |
| How can you determine pKa from a pH curve? | at the half-equivalence point pH=pKa, because half of the AH has been converted to A- and H+; so find the equivalence point, divide by two |
| What is a buffer? | A solution for which pH remains almost constant/changes only slightly for small additions of H+ or OH- |
| How would you make a buffer? | By mixing a weak acid with its conjuguate base (for example CH3COOH with CH3COO-) OR by mixing a weak acid with a base (for example CH3COOH with NaOH) |
| How do you work out the pH of a buffer given volumes and concentrations for [acid], [salt] and Ka? | work out moles of acid(AH) and salt(A-); calculate the concentrations using TOTAL volume; rearrange expression of Ka: [H+]=( Ka x [acid] )/ [salt] |
| How do you work out the pH of a buffer given volumes and concentrations for [acid], [NaOH] and Ka? | work out moles of acid(AH) and NaOH; the difference between molesAH and molesNaOH is the moles of AH unreacted; moles NaOH=moles salt;calculate the concentrations using TOTAL volume; rearrange expression of Ka: [H+]=( Ka x [acid] )/ [salt] |
| Calculate the pH of the buffer made when 0.2mole.dm^-3 of CH3COOH is mixed with 0.1mole.dm^-3 CH3COONa? (pKa=4.77)? | 4.50 |
| Calculate the pH of the solution made when 500cm3 of 1mole.dm^-3 of CH3COOH is mixed with 500cm3 of 0.4mole.dm^-3 NaOH? (Ka=6.25x10^-5)? | 4.03 |
| Explain how a buffer is able to resist a change in pH when an acid is added; HA H+ + A- | 1) the conjuguate base A- will react with the added H+ to form AH; 2) as [H+] is increased, system minimise change by decreasing [H+]/removing the added H+ and equilibrium shifts to the left |
| Explain how a buffer is able to resist a change in pH when an alkali is added; HA H+ + A- | 1) the H+ will react with the added OH-; 2) as the H+ is reacting, the system minimise change by increasing [H+] and equilibrium shifts to the right |
| Explain how buffers in the blood prevent pH from falling when extra acids are produced | H+ is removed by HCO3- to form H2CO3/carbonic acid; the carbonic acid is then converted into CO2 by an enzyme; the CO2 is exhaled |
| Calculate the pH of 0.02 mol dm-3 sulphuric acid | pH= -log (0.04) = 1.398 |
| State the two forms of the equation that link Ka and pKa | pKa = -logKa and Ka = 10^-pKa |
| Calculate the pKa of CH3COOH (Ka=1.74x10-5) | pKa=4.76 |
| Calculate the Ka of Phenol (pKa=9.89) | Ka=1.28x10-10 |
| Describe what a buffer solution is | A solution in which minimises the changes of PH on additions of small amounts of acid or alkali. |
| Give an example of a weak acid and a strong acid | A weak acid is ethanoic acid and a strong acid is hydrochloric acid |
| Write out the expression for Ka. | Ka = [H+] x [conj base] / [weak acid] |
| Calculate the pH when there is 0.35 moldm-3 of sodium hydroxide and 0.025 moldm-3 of propanoic acid when Ka is 1.45 x 10-5. | |
| Define a Bronsted-Lowry acid. | A proton donor |
| Explain what is meant by the strength of an acid. | How much an acid dissociates in aqueous solution – strong acids 100% dissociate. |
| Identify the conjugate acid-base pairs in this equilibrium:HNO3 + HNO2 NO3- + H2NO2+ | HNO3 = acid 1. HNO2 = base 2. NO3- = base 1. H2NO2+ = acid 2. |
| Write an expression for the acid dissociation constant, Ka, of phenol, given the full equation: C6H5OH + H2O <=> H30+ + C6H5O- | Ka = [H3O+][C6H5O-] / [C6H5OH] |
| How can you prove an acid is a weak acid by taking a single pH measurement? | If the single pH measurement is greater than 1, the acid is weak. |
| CHALLENGE: A weak organic acid AH has the following percentage composition by mass: C 40.0%, H 6.7%, O 53.3% and the RMM is 60.0. Calculate the empirical and molecular formula of AH. (3) | 40/12=3.33, 6.7/1=6.7, 53.3/16=3.33; /by smallest 3.33= 1, 2, 1;Empirical formula= CH2O; Mass CH2O= 12 +2 +16=30;Emp mass x2= Molecular mass; Molecular formula= C2H4O2 |
| Define pH (1) | pH=-log[H+] |
| CHALLENGE: pH of stomach is 1.3. Patient should produces 2dm^3 of gastric juice in a day. Patient is prescribed tablets containing Al(OH)3. Calculate mass of aluminium hydroxide required to raise the pH from 1.3 to 2.0. 3HCl + Al(OH)3 --> AlCl3 + 3H2O | pH2:[H+]=10^-2=0.01;pH1.3:[H+]=10^-1.3=0.05;To raise pH from 1.3 to 2.0 mean [H+] changes from 0.05 to 0.01=-0.04mole.dm-3;As 2dm^3 is made, the total amount is 2x0.04=0.08;using ratio 0.08/3mol of Al(OH)3 needed= 0.0266mol;RMM=78;mass=0.0266 x78=2.08g |
| The control of blood pH is important. This is achieved by HCO3- ions in blood plasma. Using appropriate equations, explain how these ions can act as a buffer solution. (3) | when adding extra H+= HC03- + H+ <=> CO2 + H2O, equilibrium shifts riht AND HCO3- reacts with extra H+.; when adding extra OHequilibrium shifts left (as [H+] decreases) and OH- reacts with CO2 to form HCO3- |
| What is the approximate pH at equivalence point for a weak acid-strong base titration | above 7 |
| What is the approximate pH at equivalence point for a weak base-strong acid titration | below 7 |
| What is the colour change when you add HCl onto 25cm3 of a weak base? | yellow to red |
| What is the colour change when you add NaOH onto 25cm3 of a weak acid? | colourless to pink |
| Define a Bronsted acid/base | A chemical that can donate/accept a proton |
Created by:
UrsulineChem
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