Chapter 20 Acids And Bases And Ph Flashcards
Definition for an acid and alkali
Species that dissociates in aqueous solution ti form H+ ions
Species that dissociates to form OH- ions in aqeuous solutions
Salt is an acid whose H+ ions have been replaced with metal
What is a BRONSTED LOWRY acid and base
BL acid is a species that can donate a H+ ion
BL alkali is a species that can accept a H+ ion
What are the conjugate acid base pairsm
These are species on either side that can INTERCONVERT THROUGH the donating and accepting of the H+ ions
Just identify which donate to which and label pair 1 and 2
Monobasic dibasic tribasic acid?
Refers to the total amount if H+ ions an acid can dissociate or replace per molecule , either by metal or ammonium Jon
How does an acid split in water ACTUALLY. What do we get
(What we used to
And where would we still use)
So Hcl + H20–> H3O+ cl- = HYDRONIUM ION
We normally do Hcl + Oh- —> H20 + cl-
But this is rated for aqeuous solution instead I guess
So again using hydronium how write neutralisation of acid by a base
H30+ + oH- —> 2h20
If we use H+ + Oh- calm but real way is abive
How to do ionic equations again
- what if it’s a cabrinate
Write full
Split aqeuous ones into ions
Cancel spectator
If it’s a cabrinate and it’s a solid, as it change state will STILL HAVE TO WETIE IN IONS
What are all acid reactions 5
Acid + metal = hydrogen and salt
Acid and alkali = salt and water
Acid and metal oxide = salt and water
Acid and metal hydroxide = salt and water
Acid and carbonate = salt water and co2
Main ionic equations for carbonate reactions
Acid and alkali?
H+ + co32- = h2o + co2
H+ + Oh- = water
What state are group 1 metal carbonates normally
Aqeuous so put that sign
What is idea behind pH and H+ concentration
The higher H+ concentration the lower the pH and more acidic. This differs by negative powers of 10, if I increase conc of H+ by 10 then I lower pH by 1, and want to go to Ph 4 need to dilute by 1000
Equations for pH and reverse
PH = -log [H+]
[H+] = 10 -Ph
Why is finding pH if string and weak acids different?
string acid fully dissociate and weak partially so different formula must be used
How to find pH of string acid , WHAT ASSUMPTION
Assumption is : As string acid fully dissociate, the conc of HA is the same as conc of H+ ion
So with the conc of HA do- log
Same for reverse
For strong acid what to REMEMBER IF ITS MONOBASIC OR DIBASIC
IMPORTANT
here you must multiply the conc then by 2 or 3 etc
And if reverse DIVIDE BY 2 OR 3 ETC
For seeing how concentration has changed after diluting, what formula can be used and why?
ABBAS FORMULA
We know that the MOLES is constant so c1v1 = c2v2, and rearrange
What is the acid dissociation constant and what does this value tell you
What type of acid
It’s when a WEAK ACID dissociates and you find ka kinda for it because equilibrium
- a higher Ka means higher disscoation strength as more ti right and thus more STRONG acid
Units for ka
Always moody-3 as they cancel
What do we do to make Ka values easier to compare, and what’s formula and revers
PKa = -log Ka
Ka = 10-pka
Thus will a stringer acid have higher or lower Ka and PKa?
Ka is dissociation stentgh thus stringer weaker acid will have a HIGHER KA
And since PKa is -log function, a higher Ka will give a lower PKa
So lower PKa jus like oH means STRINGER ACID
What are string acids in general and what are weak
2 new weak and what formula
String is shanc weak is carboxylic
Sulfurous H2So3
And nitrous HNo2 are also both weak
For dibasic and tribasic. Why is it that the first dissociation is much more strong in Ka values compared to second and third
Think like easy to lose a h+ on a neutral, up when charged it’s effort, so it becomes harder, so dissociates less and thus gives a higher pKa
Remember to calculate Ka like with Kc and Kp, what must the state be in (start or’
At equilibrium , and so here approximations come in
What 2 APPROXIMATIONS MUST MAKE TO SIMPLIFY KA EQUATUIN FOR A WEAK ACID
Ka = H+ x A-/ Ha all at equilibrium
1) approximation that H+ at equilibrium = A- at EQUILIBIRUM
- this is because the HA will dissociate to produce equal concentrations of both so at equilibrium this will be equal too
2) approximation that HA at equilibrium = Ha at start
- this because Ha equilibrium = Ha start - H at equilibrium
But since H+ is so small due to being weak acid, it can be negligible
New formula is Ka = H+ squared eqm/ Ha at start
Approximated formula for Ka of a weak acid?
Ka = H+ squared at eqm/ HA at start
How to find Ka experimentally then using two measurements
First is the one given, the conc of the weak acid
And second is the Ph
Then rearrange and put into formula
Further breakdown into the approximations, when does approximation one not hold
Approximation states that the dissociation fo water is negliblenand thus H+ = A- at eqm
However if Ph is greater than 6 then disscoation of water will be significant compared to disscoation fo acid
Thus it breaks down for VERY WEAK ACIDS , or very DILUTE solutijsn(because in dilute the disscoation fo water is great too)
When does appeoximation 2 break down
This approximation assumes the conc of the HA at start is biggggger than conc of H+ at eab so can ignore
However for weak acids this is true, for stringer acids then it breaks down
So when KA is bigger than 10^-2 it’s long
What is the ionic product of water kW
This is the disscoation of water
Ka = H+ x OH-/ h20
But h20 conc is constant
So simplified to K2 = H+ x OH-
And this is constants at different TEMPERTAUREs like all K values
So what is the ionic product of water at room temp 25° 298k
Kw = 1x10^-14
How can use the value of K2 to find the pH of water at a TEMPERTAURE
Water must be neutral, so it produces the same amount of H+ and Oh-
Thus kw= H+ ^2
So reraanrge fo h+
And do - log h+
At 25 ° this becomes pH of 7
What else can K w constant control
As all aqeuous soktuijsnhwge water in them kW controls the H+ and Oh- conc of these too!
Use same idea
If acid of aqueous is pH 3, then conc of Ph- will be -11 to multiply to - 14
Finally how to find thr pH of a STRING ALKALI
String alkali is one that fully dissociates to produce OH- ions
1) the conc of the Oh- is the same as the conc of the base
2) multiply this conc by molar quantity if Monobasic tribasic
Now at 25 kw is 1x10^-14
So rearrange for H+
And find pH
Summary how to find oH for strong acid weak acid and strong base
String acid = conc of acid is conc of h+ x Monobasic whatever calm
Weak acid use Ka and approximations to find H+ , rearrange
String base assumption cinc if base is conc of oh- x Monobasic whatever
Rearrange for h+ using kw
And log
Why is ph better than conc h+
H+ deals with negative indices over a wide range and lH makes it manageable to compar e
How to find the percentage of molecules dissected
Molecules of h+ dissociated = molecules of original
So do h+ / orig al x 10p
Question wrong sbout 0.1 sulfuric acid thst dissociates twice and then gives 0.96 pH? How to answe thst now
First off identify thst the first dosscistion is full, but second dissociation is partial , so the expected pH final value will change
If it dissociated completely fully, then the conc of H+ would be 0.1 x 2 (assumption thst they are the same for string acids x the DIBASIC FSCTOR)
Now - log of 0.2 gives you a pH of 0.7, which isn’t 0.96
Finally explain that as the second dissociation is only partial, the pH won’t reach as low as it does
2) OR say the actual h+ conc is 10-0.96, whereas we saying it’s 0.2, which it won’t be be sure of second dissdostion
IMPORTANT = got wrong
Student assumes the concentration of a weak acid initially is the same as at equilibrium . The measured pH is not the same as caculted, explain why
(It also has a decent high kA)
- remember the assumption here only holds for acids that are WEAK and have a low kA in General. - Remmeber the comic at equilibrium = conc at start - H+ at equilibrium. But we said weak acid = so h+ is insignificant, so it’s pretty much the dame
- however as this had a LARGE KA, we can’t use the approximation, and thus the conc at equilibrium is significantly LOWER than conc at start , so values are different
(What you said was the reason approximation one breaks down, )
Again go through why both approximations in KA calculation breaks
1) this one breaks because we ignore the disscostion of Hc in water in our equations and thus can equate H+ = A-
2) this one breaks down if weak acid is not thst weak and KA is big, because equilibrium = start - h+ at equilibrium, and for a weak acid h+ at equilibrium is really small,but as soon as kA is higher it isn’t small, so the cocnetrationsare SINGNICANTLY DIFFERANT
What value or ka does it need to be for approximation 2 to breakdown
Greater than 10.-2 and for dilute solutions
