Acids and Bases

CHM 4- acids + bases AQA

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Review from AS level

AN ACID IS A SUBSTANCE WHICH WILL RELEASE PROTONS (H+ IONS ) WHEN IT IS DISSOLVED IN WATERE.g. HCl, HNO3, H2SO4 etc.

A BASE IS A SUBSATNCE WHICH WILL REACT WITH AN ACID TO FORM A SALT (it neutralises the acid!!!)E.g metal oxides, metal hydroxides, metal carbonates ammonia

AN ALKALI IS A WATER SOLUBLE BASE WHICH RELEASES HYDROXIDE IONS (OH-) INTO SOLUTION E.g sodium hydroxide, potassium hydroxide, aqueous ammonia

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Bronsted-lowry acid base equilibria in aqueous sol

ACID----> proton donor (will release H+ ions)

BASE---->proton acceptor ( will accept H= ions)

ACID- BASE EQUILIBRIA INVOLVES THE TRANSFER OF PROTONS E.g

NH3+ H2O<----> NH4 + OH-

base <----(H+)acid-

Here water is acting as the acid as it donates a H+ ion to NH3 to form NH4+ ( which is the conjugate acid of NH3 and OH- is the conjugate base of H2O!

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definiton and determination of pH

pH----> *p= power *H= concentration of H+ ions [H+]

pH= -Log [H+]. [ ] represents the concentration in mol/dm3

To find [H+]= inverse log= 10^-pH

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pH calculations

using pH to find concentration/ using concentrations to find pH

1. find pH of- 0.0500M HNO3

  • Equation- HNO3---> H= NO3-
  • [H]= 1 x 0.0500= 0.0500M (as 1:1 ratio)
  • pH=log of H+ ( 0.0500)
  • = 1.30 ( always give pH to 2.d.p)

2. Calculate the concentration of the following STRONG acids

  • HCl with pH 1.70
  • pH = 1.70 HCl---> H+ Cl (1:1 ratio)
  • [H+]=inverse log to find pH-- 10^-1.70= 2.00x 10^-2 mol/dm3
  • HCl= H+ as 1:1 ratio so HCl= 2.00x 10^-2mol/dm3
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calculations of bases

1.Calculate the pH of the strong Base

  • 1.50 M KOH
  • KOH---> OH- + K+ 1:1ratio
  • [OH-} = 1.5M
  • [H+] =1x10^14 / 1,50= 6.67x 10^15
  • pH= -log [H+]
  • = 14.18

2.Calculate the concentration of the following base;

NaOH pH= 14.70

  • NaOH--> OH + NA+
  • [H+]= 10^14.70= 2.00 x10^-15M
  • [OH]= Kw/ [H+]= 1x10^-14 / 2.00x10^-15= 5M
  • 1:1 ratio so NaOH= 5M
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the ionic product of water

Water= AMPHIPROTIC meaning it can react with itself. Also, it is weakly dissociated!

H20 + H20---> H30+ + OH- ONE WATER WILL ACT AS AN ACID, THE OTHER, A BASE!

WE CAN SHORTEN THIS EQUATION TO: H20-<-----> H+ + OH-

THERFORE WE CAN DERIVE AN EXPRESSION FOR Kc:

Kc= [H+][OH-] / [H2O]

rearrange--- Kc x [H2O]= [H+][OH-] so few water molecules are ionized that their concentrations saty effectively constant

Kc x [H20] =will be a constant Kw----> Kw= [H+][OH-] the value of Kw is found to be 1.0 x10^-14 mol^2dm^-6

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Weak acids and Bases! Ka for weak acids

Weak acids and weak bases only dissociate slightly in aqueous solution. 'PARTIAL DISSOCIATION!' Therefore we do not know the concentration of H30+ ions.

E.g- (HA=any weak acid)

HA + H2O <----> H3O+ + A- (H2O accepts H+ ions)

Kc=[H3O+] [A-]

[HA][H2O]

for a weak acid conc. of the H2O =large & not much reacts, therefore [H2O] can be considered to be a constant!

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Ka for weak acids!

Therefore when we are calculating Ka we leave out water from the equation as it is constant!

  • Kc=[H3O+] [A-]

    [HA] =k

  • rearrange- Kc.K = [H3O+] [A-]

    [HA]

  • give Kc.K symbol 'Ka' (the acid dissociation constant)

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Ka for weak acids continue.

Ka= [H3O+] [A-} Or we can write HA<----> H+ A- therefore Ka=[H][A-]

[HA] [HA]

From this expression for Ka it is obvious that the stronger the acid the greater the value of Ka!

equations you need to know-- pKa= -logKa Ka= 10^-pKa

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calculations using Ka

1. use Ka to find the pH of a weak acid!

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calculations using ka

2. Find the concentration (or Ka) of a weak acid!

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pH curves, titrations and indicators.

(http://www.chemguide.co.uk/physical/acidbaseeqia/summary.gif)

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pH curves, titrations and indicators.

Weak acid-etahnoic strong acid- hydrochloric acid

Weak base- Ammonia Strong base- Sodium hydroxide

All the graphs ( except weak acid-weak base) on the slide before have a vertical bit--- this is the EQUIVALENCE POINT or END POINT.

EQUIVALENCE POINT = The point at which sufficient bas has been added to just neutralise the acid. (or vice verse)

END POINT= The volume of alkali or acid added when the indicator just changes colour.

HALF EQUIVALENCE= the point at which half the equivalence volume of strong base has been added to the weak acid.

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How can pH curves be used to select suitable indic

Acid/Base titrations use indicators to find the concentration of a solution of an acid or alkali

A suitable indicator for a particular titration needs the following properties

1.colour change must be sharp at the end point

2. End point given by the indicator must be the same as the equivalence point

3. Indicator must give a distinct colour change

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what indicators should be used?

1.STRONG ACID/STRONG BASE---> either Methyl orange or phneolphthalein can be used as there's a rapid pH change over the range for both indicators.

2. STRONG ACID/ WEAK ALKALI---> Only Methyl orange

3. WEAK ACID/ STRONG ALKALI---> phneolphthalein is used because the pH changes rapidly over phneolphthalein's range but not Methyl orange.

4.WEAK ACID/ WEAK ALKALI---> no indicator is suitable, you are better off using a pH meter.

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Buffer action

BUFFER= solutions which maintain an almost constant pH, even when small amounts of strong acids or bases are added or the buffer is diluted!

2 Types- 1=Acidic buffers = pH below 7

weak acid + sodium salt of SAME weak acid

2= basic buffers= pH above 7

Weak base + sodium salt of SAME weak base

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Why does dilution not cause a buffer pH to change

The value of Ka and ratio of acid:salt do not change when a buffer is diluted

[H+] does not change

pH does not change

APPLICATION (USES) OF BUFFER SOLUTIONS

  • Experiments where pH control is required
  • biological systems where pH control is essential for efficient metabolism and good health
  • commercially to prevent food deterioration due to pH changes caused by bacterial or fungal activity.
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pH calculations of acidic buffer solutions

To calculate the pH of a buffer you need to know Ka of the weak acid and the concentrations of the weak acids and its salts.

E.g contains o.4M methanoic acid ( HCOOH) and

0.6M sodium methanoate(HCOO-NA+) Ka for methanoic acid = 1.6x10^4 M. What is the pH of the Buffer?

1-Write expression HCOOH ----> H+ + HCOO-, then work out equation of Ka from this.

2.Rearrange equation of Ka, to find H+

[H+]= Ka x [HCOOH]/[HCOO-]

3. pH= -Log [H+]

4. final answer should be 3.97

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