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I keep confusing Ka and pKa , plus I can't retain the formulas very well - which I'll get to (after As mocks though *sweats* ) - generally I understand the mechanism of buffer solutions in maintaining a constant pH in tandem with Le chatelier's principle. I keep forgetting the assumptions made within the formula as well

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I have learnt the following:

pH of H2O at 25C is 7

Ka is used for weak acids

Ka = [H+][A-]/[HA] = [H+]^2/[HA]

Dissociations of water and weak acid are assumed to be too low

Greater Ka means more H+ is dissociated, pKa is lower and the weak acid is better/’stronger’

pH =  -log[H+]

pKa = -log(Ka)

SA + SB: SF at 7.5-10.5, EQ at 7

WA + SB: SF 7.5-11, EQ at 9

SA + WB: 3.5-7.5, EQ at 5

WA + WB: No SF, no good indicator

Buffer solutions: WA +CB or WB + CA

Buffer solutions ensure pH remains constant

The large reserve of salt and acid makes the change in concentration of, for example H+, negligible

pH = pKa + log([salt]/[acid])

I need to revise more on the pH graphs and values. Teaching was elaborate and informative.

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Kw
- Ionic product of water
- Kw = [H+][OH-]
- Kw = [H+]²
[H+] = square root of Kw

- pH = -log[H+]
- pOH + pH = 14

Buffer solution
- Maintains pH
- Can only handle a bit of acid/base

Ka
- Acid dissociation constant
- Ka =  [H+][A-] / [HA]
Ka = [H+]² / [HA]

- Ka directly proportional to H+ dissociation
- Higher Ka = More H+ dissociation

- pKa = -logKa
- pKa inversely proportional to Ka
- High pKa = Low Ka = Weaker acid

Equivalence point
- End of chemical reaction

End point
- Color change

pH graphs
- Strong acid + Strong base
Equivalence point = 7 (neutral)
Sharp decrease = 10.5-3.5

- Strong acid + Weak base
Equivalence point = 5
Sharp decrease = 7.5-3.5

- Weak acid + Strong base
Equivalence point = 9
Sharp decrease = 7.5-11

- Weak acid + Weak base
Equivalence point = 7 (neutral)
Sharp _____ = None

# data[].attributes.body
buffer solution

weak acid and its salt or weak base and its salt

maintains pH of a solution

Kw

ionic product of water

[H+][OH-]

[H+]²

pH

- log [H+]

pH + pOH = 14

Ka

acid dissociation constant

only for weak acids

[H+]² / [HA]

assumption

[H+] from water is ignored

dissociation of HA is too small and hence ignored

pKa

- log [Ka]

Ka↑= pKa↓

better weak acid (dissociates more)

pH graphs

strong acid + strong base

sharp fall: 11 – 3.5

equivalence point: 7

suitable indicator: bromothymol blue, phenolphthalein, methyl red

strong acid + weak base

sharp fall: 7.5 – 3.5

equivalence point: 5

suitable indicator: bromothymol blue, methyl red

weak acid + strong base

sharp fall: 11 – 7.5

equivalence point: 9

suitable indicator: bromothymol blue, phenolphthalein

weak acid + weak base

sharp fall: no sharp fall

equivalence point: 7

suitable indicator: none

equivalence point

the volume of titrant required to achieve neutralisation

end point

when the indicator changes colour during the sharp fall

# data[].attributes.body
Kw is the ionic product of water

Kw = [H+][OH-] = [H+]^2

[H+] = square root of Kw

pH = -log[H+]

pOH + pH = 14

Buffer solution: a solution that can maintain the pH when a small amount of acid or base is added

Ka is acid dissociation constant

Ka =  [H+][A-] / [HA] -> Ka = [H+]^2 / [HA]

Assumptions

Ignore concentration of H+ because Kw is relatively low

Dissociation of weak acid is too low, so we assume [HA] in equilibrium = initial [HA]

Higher Ka means H+ is dissociated more, lower Ka means less dissociation

pKa = -logKa

pKa high = weak acid, lower Ka

pKa low = stronger acid, higher Ka

Equivalence point: amount of titrant added is enough to complete the neutralisation
End point: when the indicator changes colour during the sharp fall of pH

pH graphs
1. strong acid + strong base

eq point: 7

sharp fall: 10.5-3.5

bromothymol blue, methyl red, phenolphthalein

2. strong acid + weak base

eq point: 5

sharp fall: 7.5-3.5

bromothymol blue, methyl red

3. weak acid + strong base

eq point: 9

sharp fall: 7.5-11

phenolphthalein

4. weak acid + weak base

eq point: 7

no sharp fall

no suitable indicator

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what i learnt today:

formulas:

pH = - log [H+]

pH + pOH = 14

Kw = [OH-][H+] or [H+]^2

Ka = [OH]^2 / [HA]

strong acid+strong base

eq point: 7

sharp fall: 10.5-3.5

bromothymol blue (6-7.6)

phenolphthalein (8.2-10)

methyl red  (4.2-6.3)

strong acid+weak base

eq point: 5

sharp fall: 7.5-3.5

phenolphthalein

weak acid+strong base

eq point: 9

sharp fall: 7.5-11

phenolphthalein

weak acid+weak base

eq point: 7

no sharp fall

no suitable indicator

we also learnt about the acid and base graphs, definitions of eq point and end point, and calculations regarding pH, ka and kw.

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We learned that
- smaller pKa / higher Ka = better weak acid / less dissociation
- Kw is the ionic product of water
- pH = - log [H+]
- pH + pOH = 14
- Kw = [OH-][H+] or [H+]^2
- Ka = [OH]^2 / [HA]

for the graphs:
strong acid + strong base
- eq point : 7
- sharp fall 10.5 - 3.5
- use bromothymol blue, phenolphthalein, methyl red

strong acid + weak base
- eq point : 5
- sharp fall 7.5 - 3.5
- use bromothymol blue, methyl red

weak acid + strong base
- eq point : 9
- sharp fall 7.5 - 11
- use phenolphthalein

weak acid + weak base
- eq point : 7
- no sharp fall

I need to practice drawing the graph more and memorise the sharp fall of each reaction. The class was fast, but easy to understand.

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The equations we've learnt are:
- pH=-log[H+]

- pOH + pH = 14

- Kw = [OH-][H+] / [H2O] or just [H+]^2

- Ka = [OH]^2/[HA]
[A-] = [H+]

We learnt about the general assumptions made when doing questions specifically about weak acids, which are:

1. The dissociation of weak acid is often too small, so assumed to have no dissociation.

2. The [H+] produced by H2O is so small, so it is ignored.

We learnt about the relationship between Ka + pKa and the strength of the weak acid: The smaller the value of Ka, the larger the value of pKa, the weak the acid is.

We learnt graphs, its equivalent points, ranges of their sharp falls, and how and which is the suitable indicator for the following pairs:
Weak acid + strong base
weak acid + weak base
strong acid + weak base
strong acid + strong base

I think I need to memorise the relation between Ka and pKa, and how to properly draw the graphs w/ their specific end and equivalent points.

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Sorry I had to write so that it’s easier to draw the graph and write the formulas

# data[].attributes.body
I learnt the formulas for Kw and Ka.
-Ka formula
Ignores concentration of H+ions
assume that ionisation of weak acid is small

I also learnt the different types of graphs and the definition of endpoint.

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For today’s revision , I learnt that there is a thing called a weakest weak acid. It means that it is a weak acid that has trouble dissociating and it can be determined when pKa is the highest and Ka is the lowest. Thus in exchange, the strongest weak acid is an acid which can dissociate more easily and it can be determined when pKa is lowest and Ka is highest. Other than that I revised about all the formulas and assumptions we need to remember.

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In today's session, i have learnt further about buffers solution.
I also reviewed on Ka and Kw as well as the two assumptions for Ka formula.
I've learnt the relationship of Ka and pKa in terms of the strength of weak acid. (Stronger weak acid = higher Ka, lower pKa) . Lastly, I learnt how to draw specific graphs and some basic definitions.

The lesson was fast but clear.

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I learnt how to calculate the pH and the disassociation constant for an acid. I also learnt that weaker acids will have a lower acid disassociation constant. I also learnt how to read the graphs for reactions between different combinations of acids and bases, and where their sharp fall and equivalence point will be.

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I learnt about the calculation of pH and K
>pH=-log[H+] or 14-(-log[OH-])
>Kw=[H+][OH-]
can be changed by temperature which changes pH
>Ka=[H+][A-]/[HA] or [H+]^2/[HA]
can be changed by temperature which changes pH

better weak acid has higher Ka

there are various indicators and we need to use the suitable ones for different pH

# data[].attributes.body
In today's lesson, we have learnt the following equations:
- pH = - log [H+]
- pH + pOH = 14
- Kw = [OH-][H+] or [H+]^2
- Ka = [OH]^2 / [HA]

Other than that, I learnt that Kw is the ionic product of water and also how to draw graphs, and what are their equivalent point (ep) and range of sharp falls (as shown in the picture above). It is also important to remember,

↑ Ka, ↓ pKa, less dissociation, a lower pKa indicates a stronger weak acid

↓ Ka, ↑ pKa, more dissociation, a higher pKa indicates a weaker weak acid

The equivalence point is the amount of titrant added is enough to complete the neutralisation, meanwhile, the endpoint is when indicators show colour change during the sharp fall of Ph.

For each pair, a suitable indicator can be used,

strong acid + strong base

- bromothymol blue (6-7.6)✓

- phenolphthalein (8.2-10) ✓

- methyl red  (4.2-6.3) ✓

strong acid + weak base

- bromothymol blue ✓

- phenolphthalein  ✖

- methyl red  ✓

weak acid + strong base

- bromothymol blue  ✖

- phenolphthalein ✓

- methyl red  ✖

weak acid + weak base

- no sharp fall

Overall, it was a great class, as usual, and Mr Gopi has explained everything to us clearly, so it was easy to understand his teaching. I feel like I need to practice memorizing the formulas as I am still not so familiar with them, however, everything else is good so far. Thank you sir :D

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- pH=-log[H+]

- pOH + pH = 14

- Kw = [OH-][H+]

- Ka = [OH]^2/[HA]

- Acid is stronger when Value of Ka is higher and when value of pKa is lower.

# data[].attributes.body
Bronstead-Lowry theory: Bronstead-Lowry acid is a proton donor. Bronstead-Lowry base is a proton acceptor. Why conjugate acid and conjugate base? Because it is a reversible reaction. Kw is the ionic product of water. Kw = [H+][OH-]. Kc and Kw is only dependent on temperature. Kw=1.00x10-14mol^2dm^-6. Why include log10? To change the decimal into a simpler value. Negative in-front of log to change the value of Ph to positive.pOH +pH =14. Ka is called the acid dissociation constant at 298k the value of Ka for the dissociation of ethanoic acid is 1.74x10-5moldm^-3. Ka = [H+][A-]/[HA]. Ka = [H+]^2/[HA]. More Ka means that it will be a better weak acid. Better weak acid means acid that can dissociate slightly more.

# data[].attributes.body
Acids are proton donors
Conjugate acids and bases

Kw is the ionic product of water, changes with temperature.
PH = -log[H+]
Ka = [H+]^2/[HA]
and pKa = -log10[Ka]
If Ka is high, better weak acid and low pKa

Need to revise on calculations and the graphs, acid-base indicators.

The lesson was quite fast but it is still possible to follow.

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We revised on acid-base reactions. Acid is proton donor whereas a base is a proton acceptor. We learnt also about Kw and some equations:
pH = -log[H+]
Ka = [H+]^2 /[HA]
(it is better if weak acid is high)
pKa is the opposite(better is weak acid is low)

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From this lecture, I learnt about

strong/weak acid

conjugate acid/base

how to calculate for pH, -log10[H+]

Kw = the ionic product of water = [H+][OH-]. If at 298K, [H+] and [OH-] is 1x10^-14. Therefore [OH-] is assumed to be = [H+]. So it becomes [H+]^2. Kw is affected by temperature only.

Ka for weak acid calculation. Ka is the acid dissociation constant. At 298K, value of Ka is 1.74x10^-5. Ka is affected by temperature only.

Ka = [H+][A-]/[HA]

If [H+] = [A-], Ka = [H+]^2/[HA]

pKa = -log10Ka

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Weak acid - partially dissociates
Strong acid - fully dissociates

Bronsted-Lowry Acid -> proton donor
Bronsted-Lowry Base -> proton acceptor

Kw -> ionic product of water
[H+]^2

pH -> negative logarithm to the base 10 of the hydrogen ion concentration
log10[H+]

Kw = [H+][OH-]
[H+]=Kw/[OH-]

HA (aq) <=> H+ + A-
Ka = [H+][A-] / [HA]
Ka = [H+]^2 / [HA]

Better weak acid -> Ka high, PKa low

# data[].attributes.body
Today revised  about Bronsted-lowry theory
In which:
-Acids are proton donor
-Bases are proton acceptors
-Also revised about Conjugate acids and bases

Learned about
-Kw
-How to calculate PH using  -log10^H+ and  pOH + pH = 14
-Ka and equation  ([H+]^2)/[HO]
and pKa = -log10[Ka]

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Bronsted-lowry theory

Proton donor : bronsted acid

Proton acceptor : bronsted base

Conjugate acid and conjugate base is there because there is equilibrium

Kw

Ionic product of water

Concentration of water is always constant

Its value at 298 K is 1.00 x 10^-14 mol^2dm^-6

Temperature change → change in Kw

Change in Kw → change in pH

pH

Defined as

pH = -log10[H+]

pOH + pH = 14

Ka

Acid dissociation constant

At 298 K, value of Ka for dissociation of ethanoic acid is 1.74 x 10^-5 mol dm^-3

Temperature change → change in Ka

To calculate the value of Ka, make 2 assumptions

Ignore the concentration of hydrogen ions produced by the ionisation of the water molecules present in the solution

Assume that the ionisation of the weak acid is so small that the concentration of undissociated HA molecules present at equilibrium is approximately the same as that of the

pKa = -log10[Ka]

# data[].attributes.body
Revised:
Bronsted-Lowry acid and bases.
How to calculate the pH of a substance (pH = -log([H+])

Learned:
For strong acid
Kw: Ionic product of water ([H+][OH-]/1)
[1 x 10^-14 at 298K]

For weak acid
Ka: Acid dissociation constant ( [H+] [X-] / [HX] )
If Ka is high, it is a good weak acid.

Questions involve finding the concentration of [H+] using given data to find the pH.

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today learned bout
- conjugate base/acid
- weak/strong base/acid
- Kw
- PH = -log10^H+ (and no i didnt know this before)
- pOH + pH = 14
- Ka and assumptions needed to make the equation work
- pKa lower = stronger acid

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I revised on the Brønsted acid and base, including the conjugate acid and base pairs.

I learned about:
- Kw (Ionic Product of Water){value is 1 x10^-14 at 298K }
-pH = -log10 (H+)
-pH + pOH = 14
-Ka (weak acid dissociation) {value at 1.74x10^-5 mol/dm3 at 298K for ethanoic acid}
-To count the value of Ka, we need to ignore the concentration of hydrogen ions produced by the ionization of water and we also have to assume that the ionization of weak acid is so small that the concentration of undissociated HA molecules would be the same as the original acid during equilibrium
- High Ka -> better weak acid

-PKa low -> better weak acid

I need to do more practice on the calculation and revise more on the acid-base indicators.

# data[].attributes.body
Conjugate base are proton acceptors
Conjugate acid are proton donors

Strong acids and base fully dissociates in a solution
weak acids and base partially dissociates in a solution

Kw is the ionic product of water
-Kw of pure water in 298K = 1x10^-14
thus we can assume that [H+] = [OH-] so,
Kw of water = [H+][OH-] = [H+]^2 = 1x10^-14
Kw/pH only affected by temperature

Better Weak Acid, Ka-high , P[HA]-low
P[HA] = -log10[HA]
Ka = ([H+]^2)/[HA]
Ka only affected by temperature

Indicators:
-Phenolphthalein, 8.3-10 pink colour indicates inside the range
-Methyl orange, 3.1-4.4 yellow indicates inside the range and red is the opposite
-Bromophenol Blue, 3.0-4.6 Blue indicates inside the range and yellow is the opposite

# data[].attributes.body
Kw : ionic product of water, changes with temperature.
PH = -log[H+]
Ka = [H+]^2/[HA]
Ka is high if better weak acid and pKa low

# data[].attributes.body
-We need to define conjugate acid-base pairs because there is reversible reaction.
-Kw is ionic product of water and is affected by temperature. The value is 1x10^-14
-Ka is acid disassociation constant
- pH formula : -log[H+]
- pH formula : pOH + pH = 14

# data[].attributes.body
Today i learnt about:
-Kw: Ionic product of water
-Ka: better weak acid if high
-pKa: better weak acid when low
-pH = -log10 [H+]
-Ka equation = [H+]^2 / [HA]

I need to read up more on the textbook for general definitions and learn more about the testing conditions

The teaching was easy to understand althought abit fast

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Revised on :
- bronsted lowry acid & base
- conjugate acid & base

Learnt about :
- Kw -> assumption : conc. of H+ = OH-
- Kw -> [H+]^2
- pH -> -log10 [H+]
- Ka -> assumption : conc. of [A-] = [H+]
- Ka -> [H+]^2 / [HA]
- must use suitable indicator depending on the reaction

Need to read more on :
- Calculation methods for Ka and Kw
- graphs

Teaching :
detailed and clear explanations

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Base -> Proton acceptor
Acid -> Proton donor

Kw is the ionic product of water
Kw of PURE WATER in 298K = 1x10^-14
in Kw, [H+] = [OH-] so,
Kw of water = [H+][OH-] = [H+]^2 = 1x10^-14
Kw/pH only affected by temperature

Better Weak Acid -> Ka HIGH, PKa LOW
PKa = -log10(Ka)
Ka = ([H+]^2)/[HO]
Ka only affected by temperature

Need to revise on the acid-base indicators

# data[].attributes.body
Learnt about:

Kw: ionic product of water (value is 1x10^-14 at 298K)

different temperature can give different pH

pH of pure water at room temperature is 7

acid dissociation constant, Ka (value 1.74x10^-5 at 298K)

Formula: pKa= -log10Ka

greater Ka, better weak acid

lower pKa, better weak acid

acid-base indicator and its graphs

Need to review more on the indicators range and graphs.

The lecture was clear, but the part on the graphs was too fast.

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Bases
- Strong bases vs Weak bases
- Proton acceptor

Kw : Ionic product of water
Ka - high : better weak acid
PKa - low : better weak acid

More practice :
- Read textbook for intro
- Do / re-do textbook ques to test new concept

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new formulas/

pKa = -logKa

Ka = [H+]2/[HA]