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Description
Flashcards by Toby Meredith, updated more than 1 year ago
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Created by Toby Meredith
about 6 years ago
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| Question | Answer |
| Solubility | how much of an ionic compound will dissolve in a solution e.g salt dissolving in water |
| Solubility is measured in | units: g L-1 or mol L-1 (grams per litre or mols per litre) |
| Ks the constant of the equilibrium of a saturated solution of sparingly soluble compounds has no units | |
| Saturated solution | maximum amount of solid dissolved particles |
| D | is the dissolving of a solid (solute e.g salt) in a solution (solvent e.g water) |
| Calculating Ks, given solubility of a 1:1 Of 1 mol of A+ and 1 mol of B- | Ks= [A+]x[B-] |
| Calculating Ks, given solubility of a 2:1 Of 1 mol of A+ and 2 mol of B- or vice versa | Ks= [A+]x[B-]^2 (B- is squared) expression used: 4s^3 |
| Calculating s, given Ks of a 2:1 Of 1 mol of A+ and 2 mol of B- or vice versa | use rearranging to find s |
| Calculating s, given Ks of a 1:1 Of 1 mol of A+ and 1 mol of B- or vice versa | use rearranging to find s |
| Calculating ion conc given Ks steps | 1: write equilibrium equation 2: write Ks expression 3: calculate s 4:relate ion conc to s |
| Calculating Ks from solubility given in gL-1 (grams per litre) steps | 1: convert s to molL-1 (mol per litre) 2: write equilibrium equation 3: write equilibrium expression 4: substitute s for Ks by using expression Ks=s^2 5:find the answer |
| gL-1 convert to mol L-1 | s conc in gL-1 ___________________ g molL-1 (molar mass) |
| Calculating moles | number of moles (mol) = g --------------- g mol -1 |
| calculating concentration | concentration (C) = mol ----------- V note: volume is always measured in L (conversion may be needed, divide mL by 1000) |
| calculating Ks given mass and volume steps | 1: calculate moles 2:calculate conc 3:calcuate ion conc 4: write expression for Ks |
| Common ion effect | ionic compound added to a solution that already contains the same compound (common ion effect) reduces solubility |
| Reduced solubility | means that the solid won't fully dissolve in a solution |
| Increased solubility | means that the solid will dissolve in a solution |
| Calculating solubility in solutions with common ion | |
| Common ion calculations when solutions are mixed steps | 1: calculate the new conc og conc x initial volume -------------------- total volume 2: write ks expression 3: rearrange to solve for non- common ion conc |
| Ionic product apples to solution of any solute concentration | IP is calculated the same as Ks |
| Predicting Precipitation | precipitate (ppt) forms when solutions of ionic compounds are mixed or when a solid ionic compound is added to an exisiting solution of ions |
| Determining whether a ppt will form or not steps | 1: calculate the IP 2: compare if it is greater than the value of Ks or less than IP > Ks than a ppt will form IP < Ks a ppt will not form |
| calculating ion conc to form a ppt steps | 1: Ppt forms when IP > Ks 2: concentrations of ions > Ks 3: choose on ion conc and rearrange so the expression becomes [B-] > Ks ----------- [A+] 4: solve for [B-] |
| Predicting ppt the solutions are mixed when two solutions are mixed the ion conc is diluted steps | 1: used expression: og conc x initial volume -------------------- total volume 2:calculate the conc of each ion present in the solution, separately 3: calculate IP 4: compare IP and Ks to see if a ppt forms or not |
| Predicting ppt when grams of solute (solid) are added steps | 1: find moles 2: find conc 3: find IP 4: compare IP and Ks |
| Calculating minimum quantity required for a ppt to form steps given the question "What is the minimum mass?" | 1:calculate conc of specific ion (mass wants to be found) rearrange Ks= [A+] [B-] equation to get this 2:calculate moles 3: calculate mass by rearranging moles equation |
| Factors affecting solubility | Temperature Common ions pH Complex ions |
| increased temperature, increases solubility | Increased temperature means increased solubility -more energy is available for dissolving -equilibrium shifts right to use up this energy (endothermic reaction) ---------> products are favoured |
| decreased temperature, decreases solubility | decreased temperature means decreased solubility -little energy is available for dissolving -equilibrium hardly shifts as there is not much energy |
| Common ions present, reduces solubility | Equilibrium will shift to the side where there is the most conc of ions to oppose the change (solid) hence it uses up the common ion (solid) to make more of the non-common ion (ions) making more products rather than dissolving the solid |
| added complex ions to metal cation, solubility increases | complex ion forms the solubility equilibrium shifts to dissolve the solid and replace the ions that were used to make the complex ion |
| important complex ions | |
| important complex ions | |
| low pH with high conc of H3O+ ions, increases solubilty | Strong acids completely dissociate sparingly soluble compound has an ion that will react with H3O+ ions , the equilibrium will shift to dissolve the solid to make more of the H3O+ ions. |
| high pH with high conc of OH- ions, increases solubilty | Strong bases completely dissociate sparingly soluble compound has an ion that will react with OH- ions , the equilibrium will shift to dissolve the solid to make more of the OH- ions. |
| Comparing the solubility of acidic and alkaline conditions (acid and base) explaining why something dissolves better in a solution with a very low pH and a very high pH | this is because particles of the solid fully dissociate hence the solid becomes very soluble |
| Strong Acids | -low pH -donates all protons -has a positive charge -it forms its conjugate base -completely dissociate, high conc of H3O+ ions -small pKa -large Ka |
| Weak acids | -higher pH of a strong acid -donates some protons -has a slightly positive charge -it forms its conjugate base -partially dissociates, low conc of H3O+ ions -large pKa -small Ka |
| Strong bases | -high pH -accepts all protons -has a negative charge -it forms its conjugate acid -completely dissociates, high conc of OH- ions -small pKb -Large Kb |
| Weak bases | -lower pH than strong base -accepts some protons -has a slightly negative charge -it forms its conjugate acid -partially dissociates, low conc of OH- ions -large pKb -small Kb |
| Acidic salts | some salts form weak acids when they dissolve in a solution thus making small amount of H3O+ ions and other ions that only partially dissociate |
| Basic salts | some salts form weak bases when they dissolve in a solution thus making small amount of OH- ions and other ions that only partially dissociate |
| Calculating pH of a acidic salt in solution | 1:write Ka Expression (products/ reactants) 2: Make two assumptions e.g Ka = [H3O+]^2 (conjugate base = H3O+) -------------------------------------- [weak acid] 3: substitute number values 4: rearrange to find [H3O+] 5: calculate pH |
| Finding Ka (acid dissociation constant ) | weak acid + H20 --> conjugate base + H3O+ Ka = [conjugate base][H3O+] ------------------------------------- [weak acid] Ka values are small for weak acids |
| Conjugate acid/base pair examples | Conjugate acid Conjugate base NH4+ <------> NH3 H3O+ <------> H2O |
| pKa and its formulars | |
| pKa is used to give the measure of the strength of an acid | larger pKa compared to Ka the less the solid dissociates in water thus it is a weak acid (Ka<pKa) smaller pKa compared to Ka the more solid dissociates in water thus is a strong acid (Ka>pKa) |
| Calculations of Weak Acids making assumptions | make 2 assumptions 1: no H3O+ ions comes from water so [conjugate base] and [H3O+] are equal 2: [weak acid] is unchanged as the acid only partially dissociates |
| pH of a weak acid given Ka steps | 1:write Ka Expression (products/ reactants) 2: Make two assumptions e.g Ka = [H3O+]^2 (conjugate base = H3O+) -------------------------------------- [weak acid] 3: substitute number values 4: rearrange to find [H3O+] 5: calculate pH |
| Finding pH given [H3O+] | pH= -log [H3O+] |
| finding [H3O+] given pH | [H3O+] =10^-pH |
| Concentration of a weak acid given pH steps | 1:write the Ka expression 2: make assumptions 3: rearrange formulas for to equal [weak acid] e.g [weak acid]= [H3O+]^2 ------------ Ka 4: calculate the [H3O+] 5: substitue values |
| Amphiprotic substances | can act as either a base or acid by alernating donating and accepting protons |
| Ionic product of water Kw | equilibrium constant for the dissociation of water. H2O is not included in Kw as the concentration of H20 is too high that only a small amount dissociates |
| Kw formulars | Kw= 1 x 10^-10 |
| Finding Kb (base dissociation constant) | weak base + H20 --> conjugate acid + OH- Kb = [conjugate acid ][OH-] ------------------------------------- [weak base] Kb values are small for weak bases |
| Calculating Kb given Ka | Kb= Kw(1 x 10^-10) ----------------------- Ka |
| Calculating Kb given pKa | Kb= 10^-pKa |
| Calculations of Weak Bases making assumptions | make 2 assumptions 1: no OH- ions comes from water so [conjugate acid] and [OH-] are equal 2: [weak base] is unchanged as the base only partially dissociates |
| pH of a weak base given Ka steps | 1:write Kb Expression (products/ reactants) 2: Make two assumptions e.g Kb = [OH-]^2 (conjugate acid = OH-) -------------------------------------- [weak base] 3:calculate Kb from Ka 4: substitute number values 5: rearrange to find [OH-] 6:Use Kw to find [H3O+] 5: calculate pH |
| calculating concentration of a weak base given pH | 1:write the Ka expression 2: make assumptions 3: rearrange formulas for to equal [weak base] e.g [weak base]= [OH-]^2 ------------ Kb 4:Calculate Kb from Ka 5: calculate the [H3O+] from pH 6:use Kw to calculate [OH-] 7: substitue values |
| Calculating pH of a basic salt in solution | 1:write Kb Expression (products/ reactants) 2: Make two assumptions e.g Kb = [OH-]^2 (conjugate acid = OH-) -------------------------------------- [weak base] 3:calculate Kb from Ka 4: substitute number values 5: rearrange to find [OH-] 6:Use Kw to find [H3O+] 5: calculate pH |
| Relative concentration of species | |
| Conductivity of soultions | strong bases/ strong acids have good electrical conductivity as they fully dissociate thus they have a high conc of freely moving charged particles/ ions. Whereas weak acids/weak bases have bad electrical conductivity because they partially dissociate thus they have a low conc of freely moving charged particles/ions. |
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