Acids And Bases Test Questions – Flashcards
Unlock all answers in this set
Unlock answers 62
PROTONS |
| H+ ions - Hydrogen atom has only one electron, if this is lost all that remains in a proton. In water it's bonded to at least one water molecule to form the ion H3O+ |
; ; CALCULATING THE pH OF STRONG ACIDS |
Find concentration c = n / v ;- log [H+] = pH |
; ; ;CALCULATE CONCENTRATION FROM pH |
| SHIFT log(pH) = [H+]; |
; ; DILUTING STRONG ACIDS |
Calculate number of moles of acid n = c x v Find new concentration by divinding moles with new volume c = n / v Check if the acid is H2SO4 , if so then multiply concentration by 2; Calculate pH (-log[H+]; |
; ; CALCULATING pH FOR STRONG ACIDS NEUTRALISED BY A STRONG BASE |
Calculate moles of H+ ;n = c x v; Calculate moles of OH- n = c x v Calculate moles in excess;nH+-;nOH-; Calculate concentration of [H+] c = n / v pH = -log[H+] ; |
; ; Kw |
Kw = [H+] [OH-] / [H2O]; ; H2O is omitted because it's very small. This is the equilibrium constant for water 1 x 10-14 |
CALCULATE THE pH WHEN THE CONCENTRATION OF OH- IS GIVEN |
| [H+] = Kw (1 x 10-14) / [OH-] |
CALCULATING THE pH OF STRONG ACIDS DILUTED WITH WATER |
Calculate moles of OH- n = c x v Calculate new concentration of [OH-] c = n / v Calculate [H+] using Kw: [H+] = Kw / [OH-] Calculate pH: -log[H+] |
CALCULATE pH OF WEAK ACIDS |
Rearrange the equation for Ka to make [H+][A-] = Ka [HA] [H+] = [A-] So Ka = [H+]2 = [H+][A-] = [H+]2 Calculate [H+]2 then √[H+]2 to find the answer for [H+] Calculate pH -log[H+] |
IN A STRONG ACID |
| The concentration of [H+] is equal to the concentration of acid |
WHEN DOING A CALCULATION IN REVERSE |
| First thing you do when reversed is always the last thing you do |
WEAK ACIDS AND BASES |
| Partly dissociate in aqueous solutions |
ENDOTHERMIC PROCESSES |
| There will be a higher [H+] if the temperature increases so pH decreases |
EXOTHERMIC |
| Lower concentration of H+ when temperature increases, so the pH is higher |
CONJUGATE BASE |
A substance that becomes an acid by gaining a proton e.g NH3 + HBr ↔ NH4+ + Br- NH4+ is the conjugate base |
Kw CONCENTRATION OF WATER: WHY IS H2O OMITTED FROM EQUATION? |
| The concentration of water is effectivel constant, becasue it very large so H2O can be cancelled out |
WHY IS WATER ALWAYS NEUTRAL? |
| Becase the [H+] = [OH-] So they have the same number of moles, hence the same concentration |
ESSENTIAL FEATURE OF ACID BASE REACTIONS |
| The transfer of protons |
DEFINITION OF Kw |
| The ionic product of water |
DEFINITION OF pH |
| The scale for measuring acidity and alkalinity. pH = -log [H+] |
H3O+ |
| H+ ions combine with water molecules to make the oxonium/hydronium/hydroxium ion |
H+ IONS |
| Have no electrons of thier own so it can only form a bond with another species that has a lone pair of elecetrons |
DIFFERENCE IN pH |
| Difference in one pH number is equal to a tenfold difference in [H+] |
WHY IS HCl A STRONG ACID? |
In the gas phase it is covalent In an aqueous solution it is totally ionic, so there are no molecules left so it's a strong acid |
; ; pKa |
;= -log Ka; ; This is the antilog of the acid dissociation constant : Ka;= [H+][A-] / [HA] |
; ; UNITS FOR Ka |
moldm-3;moldm-3;/;;moldm-3 =;;moldm-3 |
; ; VALUES OF pKa |
| The smaller the value of;pKa the stronger the acid |
; ; TITRATION CURVE - STRONG ACID STRONG BASE* |
; ; TITRATION CURVE - STRONG ACID WEAK BASE* |
; ; TITRATION CURVE - WEAK ACID WEAK BASE* |
; ; TITRATION CURVE - WEAK ACID STRONG BASE* |
; ; EQUIVILANCE POINT |
| Point at which sufficient base/acid has been added to neutralise the acid/base. The pH at equivilance point is not always exactly 7 |
; ; TITRATION CURVES |
| Except from weak acid and weak base, there is a large and rapid change in pH at the equivilance point even though it may not be at pH 7 |
; ; HOW TO CALCULATE CONCENTRATION USING THE EQUIVILANCE POINT |
e.g we find the equivilance point of a titration is when 25cm3 of 0.0150moldm-3 base, neutralised by 15.0cm3 acid. This shows that 25cm3 base has the same number of moles of;15.0cm3 acid.; n = c x v:so we calculate the number of moles of base; = 0.0150 x (25 x 10-3) = 3.75 x 10-4;(mol of acid and base) c = n / v : so cacid;= 3.75;x 10-4 / (15 x 10-3) = 0.025moldm-3 |
; ; INDICATORS |
| To find concentrations of a solution of acid or alkali - equivilance point has exactly the same number of moles of hydrogen ions and hydroxide ions |
; ; END POINT |
| The volume of alkali or acid added when the indicator just changes colour |
; ; SUITABLE INDICATORS |
Sharp colour change(1 drop of acid/base) to give full colour change End point of titration given by indicator ;= equivilance point; Indicator gives distinct colour change; |
; ; COMMON INDICATORS |
Universal - gradual colour change Methyl orange - changes from red to yellow at pH 3- 5 Bromophenol blue - changes from yellow to blue at pH 4-5 Methyl red - changes from red to yellow at pH 4- 6 Bromothymol blue - changes from yellow to blue at pH 6-7 Phenolphthalein - changes from colourless to pink/red at pH 10 -11 |
; ; SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON STRONG ACID STRONG BASE* |
; ; SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON WEAK ACID STONG BASE* |
; ; SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON STRONG ACID WEAK BASE* |
; ; SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON WEAK ACID WEAK BASE* |
; ; HALF NEUTRALISATION POINT |
Point half way between zero and the equivilance point; pH will change very little when we add acid/base up to this point Here we can find pKa;of a weak acid because at this point HA + OH-;; H2O + A-; At this point: [HA] = [A-;] ;so,;Ka; = [H+] [A-] / [HA]; Ka;= [H+]; So -logKa;= -log [H+] So pKa;= pH |
; ; BUFFER SOLOUTIONS |
| They resist changes of acidty and alkalinity when small amounts of acid or alkali are added to them, thier pH remains almost constant; they keep the concentration of hydrogen ions and hydroxide ions almost unchanged; based on equilibrium reaction which moves direction if either ion is added |
; ; ACIDIC BUFFERS |
Made from weak acids because dissociation of weak acids is an equilibrium reaction Definition: An acidic buffer is made from a mixture of a weak acid and a soluble salt of that acid. It will maintain a pH of below 7; |
; ; HA(aq);; H+(aq) + A-(aq) |
[H+(aq);] = [A-(aq)] It's a weak acid, so both are very small because most of the HA is undissociated |
ADDING ALKALI |
HA(aq) + OH-(aq) → H2O(aq) + A-(aq)
The OH- is removed so the pH tends to stay the same |
ADDING ALKALI |
Adding H+ shifts the equilibrium to the left, so they combine with the A- ions to make HA . However, since there is limited supplies of just A-, once no more can be combined and there is an excess in H+ ions, the pH changes. Soluble salts of HA can be added that ionise and increases the supply of A-. |
FUNCTION OF WEAK ACID IN A BUFFER |
| Acts like HA which can remove OH- |
FUNCTION OF SALT IN BUFFER |
| Source of A- ions that can remove any added H+ ions |
CHANGE IN pH |
| Buffers don't ensure NO change in pH at all. Only slight changes however. A buffer can be saturated where all avaliable HA or A-;is used up |
; ; MIXTURE OF WEAK ACID |
At half neutralisation where pH = pKa; ; neutralise half the acid to get this buffer |
; ; BASIC BUFFERS* |
| Resist change, maintian pH above 7; made from a mixture of a weak base and the salt of that base |
; ; EXAMPLE OF A BUFFER |
Blood approx 7.4 because even a small change such as 0.5 can be fatal. It's buffered by this equation H+(aq) + HCO-3(aq)↔ CO2(aq) + H2O(l) adding extra H+ shifts equilibrium to right Adding extra OH- shifts equilibrium to left |
CALCULATION OF BUFFERS |
HA(aq) ↔ H+(aq) + A-(aq) Ka = [H+(aq) ] [A-(aq] / [HA(aq)] to calculate pH of buffers |
EXAMPLE OF USING Ka = [H+(aq) ] [A-(aq] / [HA(aq)] |
Buffer contains 0.100moldm-3 ethanoic acid and 0.100moldm-3 sodium ethanoate. Ka = 1.7 x 10-5 and pKa = 4.77 for ethanoic acid. Ka = [H+(aq) ] [A-(aq] / [HA(aq)] Sodium ehtanote - fully dissociated , ethanoic acid, almost undissociated. [H+] = Ka[HA] / [A-] = (1.7 x 10-5 x 0.100) / (0.100) = 1.7 x 10-5 -log(1.7 x 10-5) = 4.77 pH = 4.77 pKa = 4.77 pH = pKa with equal conc of acid and base |
IN A BUFFER SOLUTION |
| [H+] DOESN'T EQUAL [A-] SO SIMPLIFIED EXPRESSION Ka;= [H+]2;/ [HA] IS NOT VALID |
; ; CALCULATING pH OF A WEAK ACID AND STONG BASE |
Work out moles of both acid and base; Find XS moles of base Work out concentration from the moles with the new volume Rearrange Kw for [H+] and calculate the concentration of hydrogen ions Calculate pH ; |
