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Excess Problems, solving for pH

Excess Problems, solving for pH. Chem 12 Chapter 15, Page 583-585. Calculations involving strong acids and bases. A strong, monoprotic acid dissociates in water to produce H 3 O+ and its concentration is equal to the concentration of the acid

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Excess Problems, solving for pH

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  1. Excess Problems, solving for pH Chem 12 Chapter 15, Page 583-585

  2. Calculations involving strong acids and bases • A strong, monoprotic acid dissociates in water to produce H3O+ and its concentration is equal to the concentration of the acid • A strong base dissociates in water to produce OH- and its concentration is equal to the concentration of the base

  3. We will be mixing an acid solution with a base solution. • We will calculate the pH of the solution that results after mixing. • We are not trying to find the concentration of the acid or base. There will end up being an excess of acid or base, to make the resulting solution upon mixing to be either acidic or basic.

  4. Example: • Calculate the pH of the solution that results when 25.75 mL of 0.25 M HCl(aq) is mixed with 18.75 mL of 0.31 M NaOH(aq).

  5. The steps to follow: • Look at chemical formula • Calculate the amount of each reactant using the equation: mol = mol/L x L • If there is more [H+] or [OH-] within the formula, multiply the concentration by that number • The extra reactant, is it an acid or base? So how much OH- or H+ been left behind?

  6. Calculate the concentration of excess ion by using amount of mol of excess ion divided by total solution volume (in L). • Calculate the pH, using pH = - log [H+] Note: if you were solving for OH-, then you solve first for pOH, then use 14-pOH = pH to find your final answer

  7. Answer the problem now! Step 1: the equation NaOH + HCl Step 2: mol H+ = 0.25 mol/L x 0.02575 L = 6.4 x 10-3 mol H3O+ mol of OH- = 0.31 mol/L x 0.01875 L = 5.8 x 10-3 mol OH- Step 3: This is a 1:1 mole ratio, so the smallest number given gets used up first, so the hydroxide is the limiting reagent.

  8. Step 4: How much is left? (6.4 x 10-3 mol H3O+) – (5.8 x 10-3 mol OH-) = 6.0 x 10-4 mol H3O + Step 5: Concentration = moles/ total volume = (6.0 x 10-4 mol H +) / (0.02575 L + 0.01875 L) = 0.013 M

  9. Step 6: Calculate pH pH = - log[H +] = - log [0.013 M] = 1.9

  10. Another Practice Problem • Calculate the pH of the solution that results when 26.75 mL of 0.12 M HCl(aq) is mixed with 20.36 mL of 0.31 M Mg(OH)2(aq).

  11. Follow your steps Step 1: the equation Mg(OH)2 + HCl Step 2: mol H+ = 0.12 mol/L x 0.02675 L = 0.00321 mol H+ mol of OH- = 0.31 mol/L x 0.02036 L = 0.00631 mol OH- Step 3: There are 2 [OH-] in magnesium hydroxide, so multiply by 2 = 0.00631 mol x 2 = 0.0126 mol

  12. Step 4: How much is left? 0.0126 mol – 0.00321 mol = 0.00939 mol Step 5: Concentration = moles/ total volume = (0.00939mol OH-) / (0.02675 L + 0.02036 L ) = 0.199 M Step 6: pOH = - log [0.199] = 0.70 pH = 14 – 0.70= 13.30

  13. Practice: • Calculate the pH of the solution that results when 46.85 mL of 0.15 M H2SO4(aq) is mixed with 60.00 mL of 0.18 M LiOH(aq). • Calculate the pH of the solution that results when 35.00 mL of 0.25 M RbOH(aq) is mixed with 80.00 mL of 0.23 M HClO4(aq). • Calculate the pH of the solution that results when 38.65 mL of 0.35 M H2CO3(aq) is mixed with 30.00 mL of 0.32 M Al(OH)3(aq).

  14. More practice • Page 586 # 3

  15. Step 1 2HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2H2O() • Step 2 Amount of HCl = 2.75 mol/L × 0.0319 L = 0.0877 mol • Amount of Mg(OH)2 = 0.0500 mol/L × 0.125 L = 0.00625 mol • Step 3 The reactants combine in a 2:1 ratio. The amount of HCl that will react with 0.00625 mol of Mg(OH)2 is: • Amount of HCl = 2 × 0.00625 = 0.0125 mol • The amount of acid is greater, therefore Mg(OH)2 must be the limiting reactant. • Step 4 Amount of excess HCl = 0.0877 − 0.0125 mol = 0.0752 mol • Therefore, the amount of H3O+ (aq) = 0.0752 mol • Step 5 Total volume of solution = 31.9 mL + 125 mL = 157 mL • [H3O+] = 0.0752 mol 0.157 L • = 0.479 mol/L • Step 6 pH = −log 0.479 = 0.32 • Check Your Solution • The chemical equation has a 2:1 ratio between reactants.

  16. An easier method: • Amount of HCl = 2.75 mol/L × 0.0319 L = 0.0877 mol • Amount of Mg(OH)2 = 0.0500 mol/L × 0.125 L = 0.00625 mol x 2 because there are (OH)2 so there is 0.0125 mol/L • Excess: 0.0877 - 0.0125 =0.0752 mol of acid [H3O+] • pH = - log [0.0752 mol/ (0.0319 + 0.125 L)] = 0.32

  17. Step 1 HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l) • Step 2 Amount of HBr = 3.50 mol/L × 0.0800 L = 0.280 mol • Molar mass of NaOH = 40.0 g/mol • Amount of NaOH = 4.87 g / 40.00 g/mol • =0.1218 mol • Step 3 The reactants combine in a 1:1 ratio. The amount of acid is greater, therefore • NaOH must be the limiting reactant. • Step 4 Amount of excess HBr = 0.280 − 0.122 mol = 0.158 mol • Therefore, the amount of H3O+(aq) = 0.158 mol • Step 5 [H3O+] = 0.158 mol / 0.0800 L • = 1.98 mol/L • Step 6 pH = −log [1.98 ]= 0.297

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