Molarity, Dilution, and pH

Main Idea: Solution concentrations are measured in molarity. Dilution is a useful technique for creating a new solution from a stock solution. pH is a measure of the concentration of hydronium ions in a solution.

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Molarity Review• One of the most common units of solution

concentration is molarity.• Molarity (M) is the number of moles of solute per

liter of solution.• Molarity is also known as molar concentration,

and the unit M is read as “molar.”• A liter of solution containing 1 mol of solute is a

1M solution, which is read as a “one-molar” solution.

• A liter of solution containing 0.1 mol of solute is a 0.1 M solution.

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Molarity Equation

• To calculate a solution’s molarity, you must know the volume of the solution in liters and the amount of dissolved solute in moles.

• Molarity (M) = moles of soluteliters of solution

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Molarity ExampleA 100.5-mL intravenous (IV) solution contains 5.10 g of

glucose (C6H12O6). What is the molarity of the solution? The molar mass of glucose is 180.16 g/mol.

SOLUTION:1) Calculate the number of moles of C6H12O6 by dividing

mass over molar mass = 0.0283 mol C6H12O6 2) Convert the volume of H2O to liters by dividing volume

by 1000 = 0.1005 L3) Solve for molarity by dividing moles by liters = 0.282 M

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Preparing Molar Solutions• Now that you know how to calculate the molarity of a

solution, how would you prepare one in the laboratory?

• STEP 1: Calculate the mass of the solute needed using the molarity definition and accounting for the desired concentration and volume.

• STEP 2: The mass of the solute is measured on a balance.

• STEP 3: The solute is placed in a volumetric flask of the correct volume.

• STEP 4: Distilled water is added to the flask to bring the solution level up to the calibration mark.

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http://www.ltcconline.net/stevenson/2008CHM101Fall/CHM101LectureNotes20081022.htm

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Diluting Molar Solutions• In the laboratory, you might use concentrated

solutions of standard molarities, called stock solutions.– For example, concentrated hydrochloric acid (HCl) is

12 M.• You can prepare a less-concentrated solution by

diluting the stock solution with additional solvent.– Dilution is used when a specific concentration is

needed and the starting material is already in the form of a solution (i.e., acids).

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PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?

Add water to the 3.0 M solution to lower its Add water to the 3.0 M solution to lower its concentration to 0.50 M concentration to 0.50 M

Dilute the solution!Dilute the solution!

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PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you do?do?

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you do?do?

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water But how much water do we add?do we add?

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PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you dodo??

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you dodo??

How much water is added?How much water is added?The important point is that --->The important point is that --->

moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution

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PROBLEM: You have 50.0 mL of 3.0 M NaOH and PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH and PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?you want 0.50 M NaOH. What do you do?

Amount of NaOH in original solution = Amount of NaOH in original solution = M • V = = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOHAmount of NaOH in final solution must also

= 0.15 mol NaOHVolume of final solution =Volume of final solution =(0.15 mol NaOH) / (0.50 M) = 0.30 Lor or 300 mL

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PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you do?do?

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you do?do?

Conclusion:Conclusion:

add 250 mL of add 250 mL of waterwater to 50.0 to 50.0 mL of 3.0 M mL of 3.0 M NaOH to make NaOH to make 300 mL of 0.50 300 mL of 0.50 M NaOH. M NaOH. 3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute12

A shortcutA shortcut

MM11 • V • V11 = M = M22 • V • V22

Where M represents molarity and V Where M represents molarity and V represents volume. The 1s are for represents volume. The 1s are for the stock solution and the 2s are for the stock solution and the 2s are for the solution you are trying to create.the solution you are trying to create.

Preparing Solutions by Preparing Solutions by DilutionDilution

Preparing Solutions by Preparing Solutions by DilutionDilution

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TheThe pH scalepH scale is a way of is a way of expressing the strength of expressing the strength of acids and bases. Instead of acids and bases. Instead of using very small numbers, using very small numbers, we just use the NEGATIVE we just use the NEGATIVE power of 10 on the Molarity power of 10 on the Molarity of the Hof the H++ (or OH (or OH--) ion.) ion.

Under 7 = acidUnder 7 = acid 7 = neutral 7 = neutral

Over 7 = baseOver 7 = base14

pH of Common SubstancespH of Common Substances

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Calculating the pH

pH = - log [H+](The [ ] means Molarity)

Example: If [H+] = 1 X 10-10

pH = - log 1 X 10-10

pH = - (- 10)pH = 10

Example: If [H+] = 1.8 X 10-5

pH = - log 1.8 X 10-5

pH = - (- 4.74)pH = 4.74

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pH calculations – Solving for pH calculations – Solving for H+H+pH calculations – Solving for pH calculations – Solving for H+H+

If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H++] = ???] = ???

Because pH = - log [HBecause pH = - log [H++] then] then

- pH = log [H- pH = log [H++]]

Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get

1010-pH -pH == [H[H++]][H+] = 10-3.12 = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and

then the log button17

pH calculations – Solving for H+pH calculations – Solving for H+• A solution has a pH of 8.5. What is the A solution has a pH of 8.5. What is the

Molarity of hydrogen ions in the solution?Molarity of hydrogen ions in the solution?

pH = - log [HpH = - log [H++]]

8.5 = - log [H8.5 = - log [H++]]

-8.5 = log [H-8.5 = log [H++]]

Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H++])])

1010-8.5-8.5 = [H = [H++]]

3.16 X 103.16 X 10-9-9 = [H = [H++]]

pH = - log [HpH = - log [H++]]

8.5 = - log [H8.5 = - log [H++]]

-8.5 = log [H-8.5 = log [H++]]

Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H++])])

1010-8.5-8.5 = [H = [H++]]

3.16 X 103.16 X 10-9-9 = [H = [H++]]18

More About WaterMore About WaterMore About WaterMore About WaterHH22O can function as both an ACID and a O can function as both an ACID and a

BASE.BASE.

In pure water there can be In pure water there can be AUTOIONIZATIONAUTOIONIZATION

Equilibrium constant for water = KEquilibrium constant for water = Kww

KKww = [H = [H33OO++] [OH] [OH--] =] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC

More About WaterMore About Water

KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

In a In a neutral neutral solution [Hsolution [H33OO++] = [OH] = [OH--]]

so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22

and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M

OH-

H3O+

OH-

H3O+

AutoionizationAutoionization

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pOH• Since acids and bases are opposites, pH Since acids and bases are opposites, pH

and pOH are opposites!and pOH are opposites!• pOH does not really exist, but it is pOH does not really exist, but it is

useful for changing bases to pH.useful for changing bases to pH.• pOH looks at the perspective of a basepOH looks at the perspective of a base

pOH = - log [OHpOH = - log [OH--]]Since pH and pOH are on opposite ends,Since pH and pOH are on opposite ends,

pH + pOH = 14pH + pOH = 14

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pHpH [H+][H+] [OH-][OH-] pOHpOH22

[H[H33OO++], [OH], [OH--] and pH] and pHWhat is the pH of the What is the pH of the

0.0010 M NaOH solution? 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10-3-3 M) M) pOH = - log 0.0010pOH = - log 0.0010 pOH = 3pOH = 3pH = 14 – 3 = 11pH = 14 – 3 = 11

OR KOR Kww = [H = [H33OO++] [OH] [OH--]][H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M MpH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00

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[OH[OH--]]

[H[H++]] pOHpOH

pHpH

1010 -pOH-pOH

1010 -pH-pH-Log[H

-Log[H ++]]

-Log[OH

Log[OH --]]

14 - pOH

14 - pOH

14 - pH

14 - pH

1.0 x 10

1.0 x 10-

14-14

[OH[OH

-- ]]

1.0 x 10

1.0 x 10-

14-14

[H[H++ ]]

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HOMEWORK

1) How much calcium hydroxide [Ca(OH)2], in grams, is needed to produce 1.5 L of a 0.25 M solution?

2) What volume of a 3.00M KI stock solution would you use to make 0.300 L of a 1.25 M KI solution?

3) How many mL of a 5.0 M H2SO4 stock solution would you need to prepare 100.0 mL of 0.25 M H2SO4?

4) If 0.50 L of 5.00 M stock solution is diluted to make 2.0 L of solution, how much HCl, in grams, is in the solution?

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HOMEWORK

5) Calculate the pH of solutions having the following ion concentrations at 298 K.a) [H+] = 1.0 x 10-2 M b) [H+] = 3.0 x 10-6 M

6) Calculate the pH of a solution having [OH-] = 8.2 x 10-6 M.

7) Calculate pH and pOH for an aqueous solution containing 1.0 x 10-3 mol of HCl dissolved in 5.0 L of solution.

8) Calculate the [H+] and [OH-] in a sample of seawater with a pOH = 5.60.

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