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

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Presentation on theme: "Excess Problems, solving for pH Chem 12 Chapter 15, Page 583-585."— Presentation transcript:

1 Excess Problems, solving for pH Chem 12 Chapter 15, Page

2 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 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 mL of 0.25 M HCl(aq) is mixed with mL of 0.31 M NaOH(aq).

5 The steps to follow: 1.Look at chemical formula 2.Calculate the amount of each reactant using the equation: mol = mol/L x L 3.If there is more [H+] or [OH-] within the formula, multiply the concentration by that number 4.The extra reactant, is it an acid or base? So how much OH- or H+ been left behind?

6 5.Calculate the concentration of excess ion by using amount of mol of excess ion divided by total solution volume (in L). 6.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 L = 6.4 x mol H 3 O+ mol of OH- = 0.31 mol/L x L = 5.8 x 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 mol H 3 O+) – (5.8 x mol OH-) = 6.0 x mol H 3 O + Step 5: Concentration = moles/ total volume = (6.0 x mol H +) / ( L L) = 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 mL of 0.12 M HCl(aq) is mixed with 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 L = mol H+ mol of OH- = 0.31 mol/L x L = mol OH- Step 3: There are 2 [OH-] in magnesium hydroxide, so multiply by 2 = mol x 2 = mol

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

13 Practice: 1.Calculate the pH of the solution that results when mL of 0.15 M H 2 SO 4 (aq) is mixed with mL of 0.18 M LiOH(aq). 2.Calculate the pH of the solution that results when mL of 0.25 M RbOH(aq) is mixed with mL of 0.23 M HClO 4 (aq). 3.Calculate the pH of the solution that results when mL of 0.35 M H 2 CO 3 (aq) is mixed with mL of 0.32 M Al(OH) 3 (aq).

14 More practice Page 586 # 3

15 Step 1 2HCl(aq) + Mg(OH) 2 (aq) → MgCl 2 (aq) + 2H 2 O() Step 2 Amount of HCl = 2.75 mol/L × L = mol Amount of Mg(OH) 2 = mol/L × L = mol Step 3 The reactants combine in a 2:1 ratio. The amount of HCl that will react with mol of Mg(OH) 2 is: Amount of HCl = 2 × = mol The amount of acid is greater, therefore Mg(OH) 2 must be the limiting reactant. Step 4 Amount of excess HCl = − mol = mol Therefore, the amount of H 3 O+ (aq) = mol Step 5 Total volume of solution = 31.9 mL mL = 157 mL [H3O+] = mol L = mol/L Step 6 pH = −log = 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 × L = mol Amount of Mg(OH) 2 = mol/L × L = mol x 2 because there are (OH) 2 so there is mol/L Excess: = mol of acid [H 3 O+] pH = - log [ mol/ ( L)] = 0.32

17 Step 1 HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l) Step 2 Amount of HBr = 3.50 mol/L × L = mol Molar mass of NaOH = 40.0 g/mol Amount of NaOH = 4.87 g / g/mol = 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 = − mol = mol Therefore, the amount of H3O+(aq) = mol Step 5 [H3O+] = mol / L = 1.98 mol/L Step 6 pH = −log [1.98 ]= 0.297


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