Acid/Base Practice Test Questions

Titration Calculations

• Problem 1: Determine the volume of 5.45M NaOH needed to titrate 50mL of a 6.0M H2SO4 solution to the endpoint.

  • $V<em>{NaOH} = \frac{(6.0M \ H</em>2SO_4) \times (50 \ mL)}{5.45M} = 55.05 \ mL$

• Problem 2: Find the concentration of hydrochloric acid if 15 mL of the acid is titrated with 28.5 mL of 4.0M calcium hydroxide solution.

  • Balanced equation: $2HCl + Ca(OH)<em>2 \rightarrow CaCl</em>2 + 2H_2O$
  • Moles of $Ca(OH)_2$: $(28.5 \ mL) \times (4.0 \ M) = 114 \ mmol$
  • Moles of $HCl$: $114 \ mmol \times 2 = 228 \ mmol$
  • Concentration of $HCl$: $\frac{228 \ mmol}{15 \ mL} = 15.2 \ M$

• Problem 3: Calculate the molarity of an H2SO4 solution if 40. mL of it is titrated with 50.83 mL of a 2.375 M NaOH solution.

  • Balanced equation: $H<em>2SO</em>4 + 2NaOH \rightarrow Na<em>2SO</em>4 + 2H_2O$
  • Moles of $NaOH$: $(50.83 \ mL) \times (2.375 \ M) = 120.72 \ mmol$
  • Moles of $H<em>2SO</em>4$: $\frac{120.72 \ mmol}{2} = 60.36 \ mmol$
  • Molarity of $H<em>2SO</em>4$: $\frac{60.36 \ mmol}{40. \ mL} = 1.51 \ M$

• General Equation for Acid-Base Reaction:

  • $Acid + Base \rightarrow Ionic Salt + Water$
  • Example: $HCl + NaOH \rightarrow NaCl + H_2O$

• Effect of Acid Concentration on Base Volume:

  • As acid concentration increases, the volume of base needed for neutralization also increases because more $H^+$ ions require more $OH^-$ ions to neutralize.

• Indicator Colors:

  • Phenolphthalein: Clear in acids, pink in bases.
  • Universal Indicator: Red/orange in acids, blue/purple in bases.

Standardization of NaOH Solution

• Balanced chemical reaction between NaOH and oxalic acid (H2C2O4):

  • $2NaOH + H<em>2C</em>2O<em>4 \rightarrow Na</em>2C<em>2O</em>4 + 2H_2O$

• Calculations for average standardized molarity of NaOH using the provided data:

  • Trial #1:
    • Grams of $H<em>2C</em>2O_4$: 0.67 grams
    • Volume of NaOH Used: 37.00 mL - 12.95 mL = 24.05 mL = 0.02405 L
    • Moles of $H<em>2C</em>2O_4$: $\frac{0.67 \ g}{126.00 \ g/mol} = 0.005317 \ mol$
    • Moles of NaOH: $0.005317 \ mol \times 2 = 0.010634 \ mol$
    • Molarity of NaOH: $\frac{0.010634 \ mol}{0.02405 \ L} = 0.44 \ M$
  • Trial #2:
    • Grams of $H<em>2C</em>2O_4$: 0.34 grams
    • Volume of NaOH Used: 22.50 mL - 0.48 mL = 22.02 mL = 0.02202 L
    • Moles of $H<em>2C</em>2O_4$: $\frac{0.34 \ g}{126.00 \ g/mol} = 0.002698 \ mol$
    • Moles of NaOH: $0.002698 \ mol \times 2 = 0.005396 \ mol$
    • Molarity of NaOH: $\frac{0.005396 \ mol}{0.02202 \ L} = 0.24 \ M$
  • Average Molarity of NaOH:
    • $\frac{0.44 \ M + 0.24 \ M}{2} = 0.34 \ M$

Molarity of Acetic Acid in Vinegar:

• Calculations for the average molarity (M) of acetic acid solution:
  • Trial #1:
    • Volume of NaOH Used: 45.00 mL
    • Molarity of NaOH: 0.34 M
    • Volume of Vinegar: 20. mL = 0.02 L
    • Moles of NaOH Used: $(45.00 \ mL) \times (0.34 \ M) = 15.3 \ mmol$
    • Molarity of Acetic Acid: $\frac{15.3 \ mmol}{20. \ mL} = 0.77 \ M$
  • Trial #2:
    • Volume of NaOH Used: 43.00 mL
    • Molarity of NaOH: 0.34 M
    • Volume of Vinegar: 20. mL = 0.02 L
    • Moles of NaOH Used: $(43.00 \ mL) \times (0.34 \ M) = 14.62 \ mmol$
    • Molarity of Acetic Acid: $\frac{14.62 \ mmol}{20. \ mL} = 0.73 \ M$
  • Average Molarity of Acetic Acid:
    • $\frac{0.77 \ M + 0.73 \ M}{2} = 0.75 \ M$

pH Calculations

• Fill in the missing fields in the table. You do not need to show work.

• [OH-] concentration when [H3O+] goes down:

  • a. It goes up.

• If the pH of an acid goes higher, what happens?

  • a. The [OH-] concentration goes up.
  • d. The [H+] concentration goes down.
  • e. The $H_3O^+$ concentration does down.

• If a solution has a pH of 13, the [H+] is less than the [OH-].