strong acids and bases l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Strong Acids and Bases PowerPoint Presentation
Download Presentation
Strong Acids and Bases

Loading in 2 Seconds...

play fullscreen
1 / 15

Strong Acids and Bases - PowerPoint PPT Presentation


  • 127 Views
  • Uploaded on

Strong Acids and Bases. Note. It is important that you don't confuse the words strong and weak with the terms concentrated and dilute . As you will see, the strength of an acid is related to the proportion of it that has reacted with water to produce ions.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Strong Acids and Bases' - lis


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide2
Note
  • It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.
  • As you will see, the strength of an acid is related to the proportion of it that has reacted with water to produce ions.
  • The concentration tells you about how much of the original acid is dissolved in the solution.
  • It is possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid.
acids
Acids
  • When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to produce a hydroxonium ion and a negative ion depending on what acid you are starting from.
  • In the general case . . .

HA + H2O H3O+ + A-

  • The strength of an acid is defined by the equilibrium position of the dissociation reaction shown above
strong acids
Strong Acids
  • The reaction is reversible, but with a strong acid the equilibrium lies far to the right. This means the original acid is virtually 100% dissociated(ionized)
  • For example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the reverse reaction happens that we can write:

HCl + H2O H3O+ +Cl-

strong acids5
Strong Acids
  • At any one time, virtually 100% of the hydrogen chloride will have reacted to produce hydroxonium ions and chloride ions. Hydrogen chloride is described as a strong acid.
  • A strong acid is one which is virtually 100% ionized in solution.
  • Other common strong acids include: HCl - hydrochloric acid, HNO3- nitric acid H2SO4- sulfuric acid, HBr- hydrobromic acid, HI- hydroiodic acid, and HClO4- perchloric acid
strong acid
Strong Acid

There are four ways to describe a strong acid:

  • Ka is large
  • Position of the dissociation equilibrium lies far to the right
  • The equilibrium concentration of H+ approximately equal to the original concentration of HA ([H+]≈[HA])
  • The Strength of conjugate base of a strong acid is a much weaker base than H2O
working out the ph of a strong acid
Working out the pH of a strong acid
  • With strong acids this is easy.
  • Hydrochloric acid is a strong acid - virtually 100% ionized. Each mole of HCl reacts with the water to give 1 mole of hydrogen ions and 1 mole of chloride ions
  • That means that if the concentration of the acid is 0.001M then the concentration of hydrogen ions is also 0.001M
  • So finding the pH is easy, just use your equation: pH=-log[H+]
example strong acids
Example-Strong Acids

Calculate the pH of an aqueous solution containing 2.5x10-5M of HCl.

solution
Solution

So first write the dissociation expression with HCl and water

HCl + H2O H3O+ + Cl-

Second, since we know that all the HCl will be converted to products, because HCl is a strong acid, we know that the concentrations of H3O+ and Cl- will be the same as the original acid.

All we need to do now is plug the value into the pH equation

Recall from yesterday pH=-log[H3O+]

Doing so we get

pH=-log(2.5x10-5)=4.6

strong bases
Strong Bases
  • Much the same as strong acids, strong bases dissociate 100% into the cation and OH- (hydroxide ion). The hydroxides of the Group IA and Group IIA metals usually are considered to be strong bases, they include:

Group IA-

LiOH, NaOH,KOH, RbOH and CsOH(only NaOH and KOH are common because the others are expensive to work with)

Group IIA- Ca(OH)2, Ba(OH)2, Sr(OH)2

For these soultions 2 moles of hydroxide ion is produced for every 1 mole of metal hydroxide dissolved in aqueous solution.

strong bases11
Strong Bases

Much the same as acids, there are

four ways to describe a strong base:

  • Kb is very large
  • Position of the dissociation equilibrium lies far to the right
  • Equilibrium concentration of OH- compared t original strong base is approximately equal
  • Strength of conjugate acid is much weaker acid than H2O
working out the ph of a strong base
Working out the pH of a strong base
  • With strong bases this is easy as well.
  • Sodium hydroxide is a strong base - virtually 100% ionized. Each mole of NaOH reacts with the water to give 1 mole of hydroxide ions and 1 mole of sodium ions
  • That means that if the concentration of the base is 0.001M then the concentration of hydroxide ions is also 0.001M
  • So finding the pH is easy, use the equation pH+pOH=14, find the pOH and subtract from 14 to get the pH.(You could also use the Kw expression to find the [H+] then find the pH, either way is acceptable)
example
Example

Calculate the pH of a 5.0x10-2M KOH

solution14
Solution

So first write the dissociation expression with KOH and water

KOH K+ + OH-

Second, since we know that all the KOH will be converted to products, because KOH is a strong base, we know that the concentrations of K+ and OH- will be the same as the original base

All we need to do now is find pOH and use the realtionship between pOH and pH

Recall from yesterday pOH=-log[OH-]

Doing so we get

pOH=-log(5.0x10-2)=1.30

Next subtract from 14 to get pH

pH=14-pOH=12.7

H2O

quizes
Quizes
  • Go to Mr. Richards home page on Horton website
  • Go to AP page and choose from units tab, acids and bases
  • Go to pH tutorial
  • Read over sections 4 and 6 and do quizzes