MOLARITY A measurement of the concentration of a solution

MOLARITYA measurement of the concentration of a solution

Molarity (M) is equal to the moles of solute (n) per liter of solution M = mol / L

Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water.

First calculate the moles of solute:1.5 g NaCl (1 mole NaCl) = 0.0257 moles of NaCl

58.5 g NaCl

Next convert mL to L: 0.500 L of solution

Last, plug the appropriate values into the correct variables in the equation:

M = mol/L = 0.0257 moles / 0.500 L = 0.051 mol/L

MOLARITY How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution?

, rearrange to solve for moles: moles = ML

moles= (1.25 mol / L) (0.2500 L) = .313 moles solute needed

MOLARITY M = mol / L

What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL?

First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute therefore:

47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = 0.705 g/mL density of solute

Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass. 0.705 g/mL (1 mole/ 128 g) = 0.00551 mol/mL

Next convert mL to L:

0.00551 mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M

MOLARITY & DILUTION

M1V1 = M2V2

The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged.

Since the amount or moles of solute before dilution and the moles of solute after the dilution are the same:

molb = mola

And the moles for any solution can be: moles=MV

A relationship can be established such that

MbVb = molb = mola = MaVa

Or simply : MbVb = MaVa

MOLARITY & DILUTION

Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar).

M1 = 0.05 mol/L M2 = ?

V1 = 25.0 mL V2 = 50.0 + 25.0 = 75.0 mL

M1V1 = M2V2

M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI

V2 75.0 mL

MOLARITY & DILUTIONGiven a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl?

M1 = 6.00 mol/L M2 = 0.150

V1 = ? mL V2 = 250.0 mL

M1V1 = M2V2

M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl

M1 6.00 mol/L You would need 6.25 mL of the 6.00 M HCl reagent which would be added to about 100 mL of DI water in a 250.0 mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at 250.0 mL. Mix well.

Ions in Solution• Many chemicals are ionic. When dissolved

in water they dissociate into their respective ion concentrations. [ion]=ion concentration

• Because BaCl2 is ionic and dissociates:

• BaCl2 (s) Ba+2 (aq) + 2 Cl-1

(aq)

• Therefore the [ion] may differ as shown below:

• .35M BaCl2 is really .35M = [Ba+2] and .70M = [Cl-1]

Using the Solubility Chart

• Some ions in solution are totally soluble, which means unlimited amounts can be present in solution together.

• Other ions have limited solubility together.• If ions are found on the Solubility Chart:

negative ion + positive ion NOT SOL.

means that the two are not soluble(will not dissolve) or will form a solid Precipitate (PPT) and fall to the bottom when mixed reducing or removing the ions from solution

Mixing Solutions• Mix: 200ml of .5M Pb(NO3)2 solution

with 500ml of .3M NaI solution.

Find the [ion] of all remaining ions and the mass of the precipitate after mixing the two solutions.

THINK MOLES! Must find moles first

• = .1 moles Pb+2

• = .2 moles NO3-1

• x = .15 moles NaCl• = .15 moles Na+1

• = .15 moles Cl-1

• Check Solubility Chart to see if precipitates form• Cl-1 in the presence of Pb+2 is NOT SOL. (؞PPT)• Pb+2 (aq) + 2 Cl-1

(aq) PbCl2(s) (continued)

• MAUL Chart (needed because moles are changing)• Pb+2 + 2 Cl-1

PbCl2(s)

• Continued next page

Moles Pb+2 (1) (2) Cl-1

Availiable .1 .15

Used up .075 .15

Left over .025 0

Excess chemical

Depleted

[Ion] after mixing & reacting

• /.700L=.00357M = [Pb+2]• 0 moles Cl-1/.700L = 0 M= [Cl-1] • .2moles /.700L = .286 M =[NO3

-1]

• .15 moles Na+1/.700L = .214 M = [Na+1]

• Mass of precipitate: (1 ion : 1 ppt on Used-up row)• .075moles Pb+2 make .075moles PbCl2