Titration Example Problem

a) How many moles of base was added?

b) State how many moles of acid were added initially?

c) what is the molar mass of the acid?

d) what is the value of pKa for the acid?

e) what is the pH after 120.0 mL base was added. (and of course you should be able to answer, what is the pH at any point on the curve, do this at several points for practice)

f) Would the volume to reach equivalence change if the initial volume the 10.0 g HA was dissolved in changed? (and of course you should be able to answer, what is the pH at any point on the curve, do this at several points for practice)

Suppose that 10.0g of an unknown monoprotic weak acid, HA, is dissolved in 100 mL of water. To reach the equivalence point, 100.0mL of 0.10 M NaOH was used. After the addition of 50.0 mL, the pH of the solution was found to be 4.00.

Titration Example Problem

a) How many moles of base was added?

b) State how many moles of acid were added initially? @ eq. Moles base added=moles acid initially=0.010 mol

c) what is the molar mass of the acid?

MM = =

d) what is the value of pKa for the acid?

e) what is the pH after 120.0 mL base was added. (and of course you should be able to answer, what is the pH at any point on the curve, do this at several points for practice)

f) Would the volume to reach equivalence change if the initial volume the 10.0 g HA was dissolved in changed?

Suppose that 10.0g of an unknown monoprotic weak acid, HA, is dissolved in 100 mL of water. To reach the equivalence point, 100.0mL of 0.10 M NaOH was used. After the addition of 50.0 mL, the pH of the solution was found to be 4.00.

Titration Example Problem

a) How many moles of base was added?

b) how many moles of acid were added initially? @ eq. Moles base added=moles acid initially=0.010 mol

c) what is the molar mass of the acid? MM = =

d) what is the value of pKa for the acid? pH at half equivalence= 4.00=pka

e) what is the pH after 120.0 mL base was added. (and of course you should be able to answer, what is the pH at any point on the curve, do this at several points for practice)

Excess OH- = 20mL

)f) Would the volume to reach equivalence change if the initial volume the 10.0 g HA was dissolved in changed?

NO!

Suppose that 10.0g of an unknown monoprotic weak acid, HA, is dissolved in 100 mL of water. To reach the equivalence point, 100.0mL of 0.10 M NaOH was used. After the addition of 50.0 mL, the pH of the solution was found to be 4.00.

Molar Solubility-

The amount of a compound that is soluble.

Mols/L

Calculated from Ksp

Common Ion Effect and Solubility and Kf

Just like in Keq, if you add it to a salt solution with a common ion it changes the solubility

Lowered SolubilityFill in concentration to Ksp equation

You may make an ice chart if it helps, but it is not necessary.

Increased Solubility: KfIf a solution’s ion reacts with an ion from the insoluble compound the reactions can be combined by adding reactions, and multiplying the two K together.

Increased Solubility: AcidConceptual only

AB ( s )⇄ 𝐴+¿(𝑎𝑞)+𝐵−(𝑎𝑞 )¿

Le Chatlier’s Reminder:

Common Ion vs. Increasing solubility. Le Chatelier’s Principle.

AgCl

How would you decrease the solubility of AgCl?

How many grams of AgCl dissolve in 0.5L of 0.1M NaCl?

Add a common ion: either Ag+ or Cl-

AgCl0.1M0M+x

Ksp AgCl=5x10-13

+x-x

0.1+xx

=(0.1+x)(x) 0.1x

x = M (molar solubility)

5x10-12 *(0.5L)=3.58x10-10g AgCl

Conceptual Topics: Common Ion vs. Increasing solubility.

Le Chatelier’s Principle.

𝑀𝑔𝐶𝑂3⇄𝑀𝑔2+¿+𝐶𝑂3−¿

How would you decrease the solubility of magnesium carbonate? How would you increase the solubility?

Decrease: add a common ion

Increase: add something that “removes” some product.

What reacts with CO3-? An Acid!

Common Ion Effect and Solubility

You can have two different salt solutions that are perfectly soluble, but when mixed form a precipitate

See the picture for how this happens

You can calculate if this will happenQ<K no precipitate

Q=K saturated solution

Q>K precipitate

We can see if you mix two salt solutions if a precipitate will form. This is like using Q to determine which way the reaction will go.

A 10mL solution of 1.00x10-5 M silver nitrate and a 10mL solution of 1.00x10-5 sodium chloride are mixed. Does a precipitate form?

=2.5x10-11

Q>K so there is a precipitate.

A 10mL solution of 0.100 M silver nitrate and a 10mL solution of 0.1M sodium chloride are mixed. Does a precipitate form?

Estimation: 0.1*0.1 >>than 10-13

Q>>K so there is a precipitate.

A 10mL solution of 1.00x10-6 M silver nitrate and a 10mL solution of 1.00x10-6 sodium chloride are mixed. Does a precipitate form?

=2.5x10-13

Q<K so there is NO precipitate.

Ksp AgCl=5x10-13

Balance the following redox reaction in an acidic solution. Decide which is the oxidizing agent and which is the reducing agent. Describe the flow of electrons. Is the reaction spontaneous? Find G and K.

ZnS(s) + NO3-(aq) Zn2+ (aq)+ S (s)+ NO (g)

+2 -2 -2+5 +2 0 -2+2

Balance all except O and H: Done!Use oxidation numbers to balance electronsAdd H2O to balance O Add H+ to balance Hs

2e*1=2

3e*1=3

( )*3=6

( )*2=6

3 332 2 +4H2O8H++

Oxidized: SReducing Agent: ZnSReduced: NOxidizing agent: NO3

-

Electrons are given up by S2- to the N

Balance the following redox reaction in an acidic solution. Decide which is the oxidizing agent and which is the reducing agent. Describe the flow of electrons. Is the reaction spontaneous? Find G and K.

ZnS(s) + NO3-(aq) Zn2+ (aq)+ S (s)+ NO (g)

+2 -2 -2+5 +2 0 -2+2

3 332 2 +4H2O8H++

From Reduction Table:S/S2- = -0.476NO3

-/NO = +0.96 E= 0.96 + 0.476 = 1.436 V

¿−6∗96485∗1.436=−8.31𝑥1 05 𝐽

𝐸𝑐𝑒𝑙𝑙∗𝑛0.0257

=𝑙𝑛𝐾 ⇒𝐾=𝑒( 1.436∗ 6

0.0257)=335

CellsDecide which is the oxidizing agent and which is the reducing agent for the following galvanic cell. Is the reaction spontaneous? What is the Ecell? What would happen to the mass of the anode vs the mass of the electrodes at the cathode. Describe the electron flow.

Pb2+/Pb = -0.13VAu1+/Au = +1.69 V Higher Ered=cathode

Galvanic=spontaneous: so Yes spontaneous!

Au1+ is reduced: making it the oxidizing agent

Pb is oxidized: making it the reducing agent

Ecell= 1.69+0.13 = 1.82 V

Au electrode mass increases: Pb electrode mass decreases

Electrons flow from Pb to Au