1. Strong Acids and Bases

2. 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.

3. 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 + H 2 O H 3 O + + A - The strength of an acid is defined by the equilibrium position of the dissociation reaction shown above

4. 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 + H 2 O H 3 O + +Cl -

5. 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, HNO 3 - nitric acid H 2 SO 4 - sulfuric acid, HBr- hydrobromic acid, HI- hydroiodic acid, and HClO 4 - perchloric acid

6. Strong Acid There are four ways to describe a strong acid: 1)K a is large 2)Position of the dissociation equilibrium lies far to the right 3)The equilibrium concentration of H + approximately equal to the original concentration of HA ([H + ]≈[HA]) 4)The Strength of conjugate base of a strong acid is a much weaker base than H 2 O

7. 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 + ]

8. Example-Strong Acids Calculate the pH of an aqueous solution containing 2.5x10 -5 M of HCl.

9. Solution So first write the dissociation expression with HCl and water HCl + H 2 O H 3 O + + 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 H 3 O + 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[H 3 O + ] Doing so we get pH=-log(2.5x10 -5 )=4.6

10. 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.

11. Strong Bases Much the same as acids, there are four ways to describe a strong base: 1)K b is very large 2)Position of the dissociation equilibrium lies far to the right 3)Equilibrium concentration of OH - compared t original strong base is approximately equal 4)Strength of conjugate acid is much weaker acid than H 2 O

12. 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 K w expression to find the [H + ] then find the pH, either way is acceptable)

13. Example Calculate the pH of a 5.0x10 -2 M KOH

14. 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 H2OH2O