ch. 15 ph. what is ph? ► ph is a logarithmic measurement of how acidic a solution is. ► ph =...

Ch. 15 pHCh. 15 pH

What is pH?What is pH?►pH is a logarithmic measurement of pH is a logarithmic measurement of

how acidic a solution is.how acidic a solution is.► pH = “pouvoir hydrogen” = hydrogen pH = “pouvoir hydrogen” = hydrogen

powerpower► If the pH = 1 to 6.9, it’s an acidIf the pH = 1 to 6.9, it’s an acid► If the pH = 7, it’s neutralIf the pH = 7, it’s neutral► If the pH = 7.1 to 14, it’s a baseIf the pH = 7.1 to 14, it’s a base

The pH Scale:The pH Scale:

Where pH comes from:Where pH comes from:► In water, some water molecules split up In water, some water molecules split up

into Hinto H++ and OH and OH-- ions like this: ions like this:►HH22OO(l) (l) ⇆ H⇆ H++

(aq) (aq) + OH + OH--(aq)(aq)

►HH++ could also be H could also be H33OO++ or just a proton... or just a proton... ► In pure water, the reverse reaction is In pure water, the reverse reaction is

favored (a lot!) in equilibrium:favored (a lot!) in equilibrium: [ ] means [ ] means Molarity or concentrationMolarity or concentration

►[H[H++] = 1.0 x 10] = 1.0 x 10-7-7 moles/L moles/L►[OH[OH--] = 1.0 x 10] = 1.0 x 10-7-7 moles/L moles/L

AcidAcid Neutral Neutral Base Base

Fun with Fun with LOGS

LogarithmLogarithmA logarithm is an EXPONENT!!!A logarithm is an EXPONENT!!!

Definition:Definition:- exponent expressing the power to - exponent expressing the power to which a fixed number must be raised in which a fixed number must be raised in order to get a given numberorder to get a given number- a.k.a log- a.k.a log

The Relationshiploglogbb(x) = y which is the same as x (x) = y which is the same as x

= b= byy

Stated: log-base-b of y equals x

Why are we learning about Why are we learning about logs?logs?

1)1) pH is logrithmicpH is logrithmic

The exponent on the 10 is The exponent on the 10 is the pH!the pH!

►We state the concentration of HWe state the concentration of H++ ions in ions in water as its "pH" which is based on the water as its "pH" which is based on the exponent on the 10. exponent on the 10.

► In pure water, In pure water, [H[H++] = 1.0 x 10] = 1.0 x 10-7-7 moles/L moles/L►pH = -log pH = -log [H[H++] ] ►So the pH of pure water is pH = 7.0So the pH of pure water is pH = 7.0►Remember logs? ie: 10Remember logs? ie: 1033 = 1000 = 1000►Then log(1000) = 3Then log(1000) = 3

How many legs does the elephant How many legs does the elephant have?have?

What’s the pH of 0.005 M What’s the pH of 0.005 M HNOHNO33??

►HNOHNO33 H H++ + NO + NO33-- (complete dissociation)(complete dissociation)

►So, [HSo, [H++] = 0.005 M] = 0.005 M►pH = -log[HpH = -log[H++] = -log[0.005] = 2.3 ] = -log[0.005] = 2.3 (acid range)(acid range)

pH + pOH = 14pH + pOH = 14►Or: [HOr: [H++]x]x[OH[OH--] = 1.0 x 10] = 1.0 x 10-14-14

►For a base, pOH = -log For a base, pOH = -log [OH[OH--] ] ►Ex: What’s the pH of 0.002 M NaOH?Ex: What’s the pH of 0.002 M NaOH?►First, the [OHFirst, the [OH--] = 0.002 M] = 0.002 M►Then, pOH = -log[0.002] = 2.7 (try it!)Then, pOH = -log[0.002] = 2.7 (try it!)►Then pH = 14 – pOH = 11.3 (basic Then pH = 14 – pOH = 11.3 (basic

range)range)

Find the pH of the following:Find the pH of the following:► 0.025 M HNO0.025 M HNO33

► Acid, [HAcid, [H++] = 0.025; pH = 1.6] = 0.025; pH = 1.6► 0.00056 M HBr0.00056 M HBr► Acid, [HAcid, [H++] = 0.00056; pH = 3.25] = 0.00056; pH = 3.25► 0.0301 M KOH0.0301 M KOH► Base, [OHBase, [OH--] = 0.0301; pOH = 1.5, pH = 12.5] = 0.0301; pOH = 1.5, pH = 12.5► 0.0030 M Sr(OH)0.0030 M Sr(OH)2 2

►Note! Sr(OH)Note! Sr(OH)22 Sr Sr2+ 2+ + 2OH+ 2OH¯̄ (1 to 2 Ratio) (1 to 2 Ratio)► Base, [OHBase, [OH--] = 0.0060; pOH = 2.2, pH = 11.8] = 0.0060; pOH = 2.2, pH = 11.8► 0.0044 M H0.0044 M H22SOSO44

► Acid, [HAcid, [H++] = 0.0044; pH = 2.35] = 0.0044; pH = 2.35

Count Up the Black Dots!Count Up the Black Dots!

Titration:Titration:►A way to figure out what A way to figure out what

the molarity of an acid the molarity of an acid is by adding a known is by adding a known molarity base to it until molarity base to it until it is neutralized.it is neutralized.

►Or Visa Versa: Add a Or Visa Versa: Add a known acid to an known acid to an unknown molarity base.unknown molarity base.

The pH of some common The pH of some common things.things.

How to do a titration:How to do a titration:►1. Get a few mL of the unknown solution and 1. Get a few mL of the unknown solution and

add some water and a little bit of an add some water and a little bit of an indicator.indicator.

►2. Fill your buret with the “standard” acid or 2. Fill your buret with the “standard” acid or base solution; the one you know the molarity base solution; the one you know the molarity of. of.

►3. Slowly dribble from your buret into the 3. Slowly dribble from your buret into the unknown solution until the unknown solution until the endpointendpoint is is reached (shown by the indicator color). reached (shown by the indicator color). Record the mL of how much you added.Record the mL of how much you added.

How a Titration works:How a Titration works:

Titration in action, continued:Titration in action, continued:

Before the HCl is added, the Before the HCl is added, the three things in the solution are three things in the solution are water molecules, OHwater molecules, OH-- ions and ions and

NaNa+ + ions.ions.

Adding HCl is really adding HAdding HCl is really adding H+ +

ions which react with the OHions which react with the OH-- ions to make water. The ions to make water. The green spheres are Clgreen spheres are Cl- - ions ions

which don’t react.which don’t react.

At the endpoint, there are just At the endpoint, there are just ClCl-- ions, Na ions, Na++ ions, and H ions, and H22O O molecules in the solution.molecules in the solution.

Example:Example:►55.0 mL of 0.5 M NaOH are needed to 55.0 mL of 0.5 M NaOH are needed to

titrate 20.0 mL of an unknown M HCl titrate 20.0 mL of an unknown M HCl solution to the endpoint. What’s the solution to the endpoint. What’s the Molarity of the HCl?Molarity of the HCl?

►Step 1: Write and balance the reactionStep 1: Write and balance the reaction►Step 2: Use DA to find moles of HClStep 2: Use DA to find moles of HCl►Step 3: Divide moles/Liters to get the Step 3: Divide moles/Liters to get the

MolarityMolarity

Step 1: Write the ReactionStep 1: Write the Reaction►HCl + NaOH HCl + NaOH NaCl + H NaCl + H22OO

Step 2: Find the moles of HCl that Step 2: Find the moles of HCl that were in your unknown M HCl were in your unknown M HCl

solution.solution.►The “arrow” always starts with the The “arrow” always starts with the

volume of the standard solution; the volume of the standard solution; the one which came from the buret.one which came from the buret.

►0.055 L NaOH 0.055 L NaOH ? mols HCl ? mols HCl

HCl mol 0.028NaOH mol 1HCl mol 1

NaOH L 1NaOH mol 0.5NaOH L 0.055

Step 3: Divide the moles of HCl Step 3: Divide the moles of HCl by the Liters of HCl to find the by the Liters of HCl to find the

Molarity.Molarity.►We found that we had 0.028 mols HCl We found that we had 0.028 mols HCl

in our 0.020 L of HCl solution that we in our 0.020 L of HCl solution that we started with:started with:

HCl M 4.1Soln L 0.020

HCl mols 028.0Soln L

HCl molsM

A pH graph of this titration.A pH graph of this titration.

What acid rain did to ol’ What acid rain did to ol’ George…George…

►Marble is made of calcium carbonate, Marble is made of calcium carbonate, a weak base which reacts with the a weak base which reacts with the acid rain.acid rain.

Sample Problem:Sample Problem:►1. You titrate 20 mL of an unknown M 1. You titrate 20 mL of an unknown M

sulfuric acid. 15.5 mL of 1.5 M NaOH is sulfuric acid. 15.5 mL of 1.5 M NaOH is required to reach the endpoint. What’s required to reach the endpoint. What’s the concentration of the sulfuric acid?the concentration of the sulfuric acid?

►Follow the last example in the notes. Follow the last example in the notes. Write the reaction, then find the moles of Write the reaction, then find the moles of sulfuric acid with DA, then find the sulfuric acid with DA, then find the Molarity of it.Molarity of it.

►You should have gotten 0.0116 mols of You should have gotten 0.0116 mols of HH22SOSO44 , giving [H , giving [H22SOSO44] = 0.58 M] = 0.58 M

That’s All For Now!That’s All For Now!