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**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 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 solute liters of solution

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Molarity Example A mL intravenous (IV) solution contains 5.10 g of glucose (C6H12O6). What is the molarity of the solution? The molar mass of glucose is g/mol. SOLUTION: Calculate the number of moles of C6H12O6 by dividing mass over molar mass = mol C6H12O6 Convert the volume of H2O to liters by dividing volume by 1000 = L Solve for molarity by dividing moles by liters = M

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Example 2 A 350 mL solution of sodium chloride contains 17.5 g of sodium chloride. What is the molarity of this solution?

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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**

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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 NaOH and you want 0. 50 M NaOH**

PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!

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

PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add?

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**The important point is that --->**

PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution

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**Amount of NaOH in original solution = M • V = **

PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M • V = (3.0 mol/L)(0.050 L) = mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH) / (0.50 M) = 0.30 L or mL

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**PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH**

PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

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**Preparing Solutions by Dilution**

A shortcut M1 • V1 = M2 • V2 Where M represents molarity and V represents volume. The 1s are for the stock solution and the 2s are for the solution you are trying to create.

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Example How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl solution? 2 M x V1= .15 M x .250 L .0375/2.0 = V1 V1 = L V1 = 19 mL

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Example 2 What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL of 4.2 M KOH? 4.2 M x .045 L = M2 x .295 L .189/.295 = M2 M2 = .64 M

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**Letters letters everywhere**

pH, [H+], pOH, [OH-] Letters letters everywhere

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**The pH scale is a way of expressing the strength of acids and bases**

The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H+ (or OH-) ion. Under 7 = acid = neutral Over 7 = base

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**pH of Common Substances**

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**(The [ ] means Molarity)**

Calculating the pH pH = - log [H+] (The [ ] means Molarity) Example: If [H+] = 1 X 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 H+**

If the pH of Coke is 3.12, [H+] = ??? Because pH = - log [H+] then - pH = log [H+] Take antilog (10x) of both sides and get 10-pH = [H+] [H+] = = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and then the log button

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**pH calculations – Solving for H+**

A solution has a pH of What is the Molarity of hydrogen ions in the solution? pH = - log [H+] 8.5 = - log [H+] -8.5 = log [H+] Antilog -8.5 = antilog (log [H+]) = [H+] 3.16 X 10-9 = [H+]

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**More About Water H2O can function as both an ACID and a BASE.**

In pure water there can be AUTOIONIZATION Equilibrium constant for water = Kw Kw = [H3O+] [OH-] = x at 25 oC

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**More About Water Autoionization**

Kw = [H3O+] [OH-] = x at 25 oC In a neutral solution [H3O+] = [OH-] so Kw = [H3O+]2 = [OH-]2 and so [H3O+] = [OH-] = 1.00 x 10-7 M

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**pOH Since acids and bases are opposites, pH and pOH are opposites!**

pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a base pOH = - log [OH-] Since pH and pOH are on opposite ends, pH + pOH = 14

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pH [H+] [OH-] pOH

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**[H3O+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution?**

[OH-] = (or 1.0 X 10-3 M) pOH = - log pOH = 3 pH = 14 – 3 = 11 OR Kw = [H3O+] [OH-] [H3O+] = 1.0 x M pH = - log (1.0 x 10-11) = 11.00

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**[OH-] [H+] pOH pH 1.0 x 10-14 [OH-] 10-pOH 1.0 x 10-14 [H+] -Log[OH-]**

-Log[H+] 14 - pH pH

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HOMEWORK How much calcium hydroxide [Ca(OH)2], in grams, is needed to produce 1.5 L of a 0.25 M solution? What volume of a 3.00M KI stock solution would you use to make L of a 1.25 M KI solution? How many mL of a 5.0 M H2SO4 stock solution would you need to prepare mL of 0.25 M H2SO4? 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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