**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.

**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.

**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

**Molarity Example** A 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: • Calculate the number of moles of C6H12O6 by dividing mass over molar mass = 0.0283 mol C6H12O6 • Convert the volume of H2O to liters by dividing volume by 1000 = 0.1005 L • Solve for molarity by dividing moles by liters = 0.282 M

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

**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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**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).

**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!

**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?

**moles of NaOH in ORIGINAL solution = ** moles of NaOH in FINAL solution 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 --->

**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) = 0.15 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 300 mL

**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 waterto 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

**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.

**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 = .01875 L • V1 = 19 mL

**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

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

**ThepH scaleis 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 7 = neutralOver 7 = base

**pH of Common Substances**

**Calculating the pH** pH = - log [H+] (The [ ] means Molarity) Example: If [H+] = 1 X 10-10pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example: If [H+] = 1.8 X 10-5pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

**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+] = 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 button

**pH calculations – Solving for H+** • A solution has a pH of 8.5. 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+]) 10-8.5 = [H+] 3.16 X 10-9 = [H+]

**More About Water** H2O can function as both an ACID and a BASE. In pure water there can beAUTOIONIZATION Equilibrium constant for water = Kw Kw = [H3O+] [OH-] =1.00 x 10-14at 25 oC

**More About Water** Autoionization Kw = [H3O+] [OH-] = 1.00 x 10-14 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

**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

**pH** [H+] [OH-] pOH

**[H3O+], [OH-] and pH** What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10-3 M) pOH = - log 0.0010 pOH = 3 pH = 14 – 3 = 11 OR Kw = [H3O+] [OH-] [H3O+] = 1.0 x 10-11 M pH = - log (1.0 x 10-11) = 11.00

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

**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 0.300 L of a 1.25 M KI solution? • How many mL of a 5.0 M H2SO4 stock solution would you need to prepare 100.0 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?

**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.