Chapter 20 Acids And Bases And Ph

Question 1

Definition for an acid and alkali

Answer 1

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

Question 2

What is a BRONSTED LOWRY acid and base

Answer 2

BL acid is a species that can donate a H+ ion
BL alkali is a species that can accept a H+ ion

Question 3

What are the conjugate acid base pairsm

Answer 3

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

Question 4

Monobasic dibasic tribasic acid?

Answer 4

Refers to the total amount if H+ ions an acid can dissociate or replace per molecule , either by metal or ammonium Jon

Question 5

How does an acid split in water ACTUALLY. What do we get

(What we used to
And where would we still use)

Answer 5

So Hcl + H20–> H3O+ cl- = HYDRONIUM ION

We normally do Hcl + Oh- —> H20 + cl-
But this is rated for aqeuous solution instead I guess

Question 6

So again using hydronium how write neutralisation of acid by a base

Answer 6

H30+ + oH- —> 2h20
If we use H+ + Oh- calm but real way is abive

Question 7

How to do ionic equations again

• what if it’s a cabrinate

Answer 7

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

Question 8

What are all acid reactions 5

Answer 8

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

Question 9

Main ionic equations for carbonate reactions

Acid and alkali?

Answer 9

H+ + co32- = h2o + co2

H+ + Oh- = water

Question 10

What state are group 1 metal carbonates normally

Answer 10

Aqeuous so put that sign

Question 11

What is idea behind pH and H+ concentration

Answer 11

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

Question 12

Equations for pH and reverse

Answer 12

PH = -log [H+]
[H+] = 10 -Ph

Question 13

Why is finding pH if string and weak acids different?

Answer 13

string acid fully dissociate and weak partially so different formula must be used

Question 14

How to find pH of string acid , WHAT ASSUMPTION

Answer 14

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

Question 15

For strong acid what to REMEMBER IF ITS MONOBASIC OR DIBASIC
IMPORTANT

Answer 15

here you must multiply the conc then by 2 or 3 etc

And if reverse DIVIDE BY 2 OR 3 ETC

Question 16

For seeing how concentration has changed after diluting, what formula can be used and why?
ABBAS FORMULA

Answer 16

We know that the MOLES is constant so c1v1 = c2v2, and rearrange

Question 17

What is the acid dissociation constant and what does this value tell you
What type of acid

Answer 17

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

Question 18

Units for ka

Answer 18

Always moody-3 as they cancel

Question 19

What do we do to make Ka values easier to compare, and what’s formula and revers

Answer 19

PKa = -log Ka
Ka = 10-pka

Question 20

Thus will a stringer acid have higher or lower Ka and PKa?

Answer 20

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

Question 21

What are string acids in general and what are weak
2 new weak and what formula

Answer 21

String is shanc weak is carboxylic

Sulfurous H2So3
And nitrous HNo2 are also both weak

Question 22

For dibasic and tribasic. Why is it that the first dissociation is much more strong in Ka values compared to second and third

Answer 22

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

Question 23

Remember to calculate Ka like with Kc and Kp, what must the state be in (start or’

Answer 23

At equilibrium , and so here approximations come in

Question 24

What 2 APPROXIMATIONS MUST MAKE TO SIMPLIFY KA EQUATUIN FOR A WEAK ACID

Answer 24

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

Question 25

Approximated formula for Ka of a weak acid?

Answer 25

Ka = H+ squared at eqm/ HA at start

Question 26

How to find Ka experimentally then using two measurements

Answer 26

First is the one given, the conc of the weak acid And second is the Ph Then rearrange and put into formula

Question 27

Further breakdown into the approximations, when does approximation one not hold

Answer 27

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)

Question 28

When does appeoximation 2 break down

Answer 28

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

Question 29

What is the ionic product of water kW

Answer 29

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

Question 30

So what is the ionic product of water at room temp 25° 298k

Answer 30

Kw = 1x10^-14

Question 31

How can use the value of K2 to find the pH of water at a TEMPERTAURE

Answer 31

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

Question 32

What else can K w constant control

Answer 32

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

Question 33

Finally how to find thr pH of a STRING ALKALI

Answer 33

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

Question 34

Summary how to find oH for strong acid weak acid and strong base

Answer 34

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

Question 35

Why is ph better than conc h+

Answer 35

H+ deals with negative indices over a wide range and lH makes it manageable to compar e

Question 36

How to find the percentage of molecules dissected

Answer 36

Molecules of h+ dissociated = molecules of original So do h+ / orig al x 10p

Question 37

Question wrong sbout 0.1 sulfuric acid thst dissociates twice and then gives 0.96 pH? How to answe thst now

Answer 37

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

Question 38

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)

Answer 38

- 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, )

Question 39

Again go through why both approximations in KA calculation breaks

Answer 39

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

Question 40

What value or ka does it need to be for approximation 2 to breakdown

Answer 40

Greater than 10.-2 and for dilute solutions