Choater 20

Question 1

Brindled lowery acids and bases?

Answer 1

BL ACID is a species that donates a H+ ion

BL base is a species that can ACCEPT A h+ ion

So even in those nucleophikic RESCTION acids, even tho naught be electiphile NUCLEOPHILIC, if they accept or donate they could be acting as acid or base

SO WAYCH OTU

Question 2

What’s a conjugate acid base pair?

Answer 2

A pair of species that from the inter conversion of a hydrogen ion can become each other, needs BL base and acid

Always pairs

Question 3

Monobasic dibasic etc?

Answer 3

Refers to the number of Hydogen ions a species can donate in total
H2so4 is dibasic

Question 4

How does something actually neutralise IN WATER
What’s this ion

Answer 4

Instead of H+ and OH-, the H+ goes to H2o to become H3+O, hydronium ion

Here we nwomally do Hcl + OH becomes water and Cl-
But when usign aqueous soltuon do Hcl + H2O becomes H39+ and cl-

Int his case rhe H3O is BL acid now

Question 5

How ti write ionic equations, with the exception of a carbonate

Answer 5

Here write split into ions that are aqueous, and then cancel

However group 2 metal carbonates are normally solid but stil, in this case solitaire the, up

Question 6

Acid reactions

Answer 6

Acid + metal = hydrogen and salt
Acid + alklai = salt + water
Acid +. Teal oxide = salt + water
Acid + metal hydroxide = salt and more water
Acid + metal carbonate = salt water co2

Group one metals normally aqeuous

Question 7

Ph

Answer 7

Higher conc of H+ determines pH lower
Ph = - log (h+) conc
H+ 10-ph

Question 8

How ti find Ph for string acids and assumption

What to remember !

Answer 8

Assumptions is as it’s a strong acid all the acid dissociates into h+ ions

To find the cinc of H+ do conc of acid x the fact it’s mono or dibasic

Question 9

Remember how to do dilution questions, Abbas formula

Answer 9

Moles constant

Upon dilution, moles ar the same

So c1v1 = c2v2

Question 10

WHY DO WE USE LOGARITIHMIC SCALE

Answer 10

I think because we are dealing with EGATIVE INDIDICES, it’s hard to make comparisons. E tween

Therefore we do PKa and Ph

Question 11

Eva for weak acids

What does Ka and PKa mean in terms of stentgh of the acid

Answer 11

Weak acids form an EAUILOBRIA as they partially dissociate, therefore we can use an ewuaikoborum constsnt which is KA

Ka is small so we use PKa to compare

The higher the Ka, the STRINGER the weaker acid as more of it disscoates
Therefore the LOWER THE PKA adjust like with pH, the stinger the weaker acid too

Question 12

What are string acids and weak acids in general

Answer 12

Strongest acids are the shanc

Weakest acids are the cabroylic acids

But still the stringent weakest acid, able to react with carbonate a weak alklai to produce bubbles, but something like phenol can’t

Question 13

Why is the first disscoation for a weak acid always result in a lower PKa so stinger acid then second isisscistij?

Answer 13

First dissocsiton you’re removing h+ from neutral calm

Second it’s from h+ from an already negative intermediate which is harder, and thus happens less often

Question 14

Why are there Ka approximations?

Answer 14

Because we can only uses these things once Moles at EQUILBKRUM are found, but because some are very similar to each other we can make approxmstims

Question 15

What is the Ka equations and what 2 APPROXIMATIONS CAN WE MAKE AND WHY

Answer 15

Ka equatuins is the H+ and A- at equilibrium all / HA at equkibkrum

But we can write Ha at EQUILBKRUM as Ha at start - moles of H+ equkibkrum as this is equal to the change

1) okay so initially , assuming the concentration of H+ from the AQEUOUS WATER IS SMALL, we know that the dissocsiton fi HA will produce H+ and A- in EQUAL QUANTITIES.
= therefore top = H+ 2

2) and we rewrote HA at eqbm to be HA start - H+ at equilibrium
Well it is a weak acid, with a low Ka, the moles of H+ at EQUILBKRUM will be very little compared to moles of the Acid at the start which were j disscoates

So we can IGNORE the H+ at EQUILBKRUM and therefore HA start = HA at EQUILBKRUM

And final equation is KA is riougjly H+ 2 / HA at start for weak acis

Question 16

When does approximation break down and why?

Answer 16

If the KA IS ACTUALLY BIG
- then the H+ at EQUILBKRUM is no longer negligible, and will have an effect
- if KA is big it mean conc of H+ at EQUILBKRUM is quite decent and so appeimxstiojnbreakdneownn, not justified for STRONGER WEAKER ACIDS

1) for approximation number one assuming the conc of H+ due ti water is very negligible
Now this is true at room temp, where Ph is 7, and conc of h+ due ti water by Kw is 1x 10-7
But when pH is less then 7, now the conc of H+ slowly increases, and this happens at HIGHER TEMPS

Or at DILUTE AFIDS WEAK ACIDS IT BREAK DOWNS

Question 17

Again approximation 1 and 2 when break down

Answer 17

1 breaks down when very dilute (content of water big) or the pH of water is less than 7 due to temp, as then the contribution of H+ form water actually takes an effect. Doesn’t work for very dilute acids

2 breaks downs when the acid is string, as the KA will be higher and thus moles of H+ at EQUILBKRUM high too, which takes an effect in equation for moles of HA at EQUILBKRUM = HA start £ H+ at equkibkrum

Thus breaks down for high KA acids

Question 18

Ionic product of water?

How ti find neutral point of water the pH based on info

Answer 18

This is a constant at different tempertaures and controls the Ph of WATEE

At room temp it’s 10-14

We know that at neutral the conc o H+ must be the same as OH-

So we can just say H+2, which = ionic product of water

Rearrange to find H+

You’ll see you get pH 7

But at different temps the pH if neutral water changes as ionic product changes too

Question 19

How is ionic product of water useful for us in acids alklai etc

Answer 19

Every AQEUOUS SOLUTON THERE WILL BE A NUMBER OF OH - IONS AND H+

If acidic then H+ > Oh-, if ALKLI then less if neutral then equal

Therefore, if we know the temp is room temp, then we can use KW as a constant to find out how much OH- ions there is or H+ if we know one other value

Acidic solutions still contain oH- ions, if pH is 3 at temo 25, then we can figure out how much Pah - there’s we

Must multiply and thus add to give -13

Question 20

Like string acid what is the appeoximation for strong based

Answer 20

A string base is something that comoeltley dissociates ti form Oh- ions in solution

Question 21

Therefore how would we find oH if a string base

HWAT TI REMMEBER

Answer 21

oh if a string are, well we can find the Oh cinc first

This is = to conc if base at start x mono / di basic type base

Now we have Oh cinc, use i2 to find H+ conc

And then fidnpah

Question 22

Why use ph and PKa instead of H+

Answer 22

H+ deals with numbers if negative indicies over a wider range, so using ph allows us to use numbers which are much more comparable

Question 23

How ti find PERCENTAGE OF MOLECULES DISSOCSTED

Answer 23

Use conc Of H+ / CONC OF ORIGNAL TO GIVE INDICATION HOW MNAY MOLECULES DISSOCed

Question 24

Why does 0,1 sukfruic acid that disscoates twice, fire tfukly snd then partially give a pH of 0,96

Answer 24

Well if it dislodged fully it would give ph of 0.7 usign dibasic rule

The fact that first disscoation is full but second is partial. Meaning not enough h+ ions will be released in order to reach a ph that low

And thr conc of h+ they say. Well is 0.1 which is not same as 0.2, and this is because if the little disscoation from second ssicostion

Question 25

Finally go through H+ aprlidmsitins snd why breaks don

Answer 25

1) apoeidmstes h+ from water small enough so we can say h+ = A-
Thing is when very dilute, or oH less than 6 , the ph of water is quite high so appeoximation breakdown

2) appeimxsted the H+ of euaonorum negligee cimaored to Ndowcistrd conc of aha at start as it’s a weak acid, so can just say ha at daub = ha start

Problem with this if the Ka is big then the h + at eqm will Acc bue quire a lot, so the allridmsitoj breaks down and can’t use

So doesn’t work for stringer weaker acid