Strong Acid-Strong Base Mixture Calculations

Example of Mixing Strong Acid with Strong Base

Overview

  • This section covers a problem involving the mixing of a strong acid and a strong base, specifically calculating the pH of the resulting mixture.
  • Given Data:
    • Volume of strontium hydroxide solution: 150 mL
    • Concentration of strontium hydroxide (Sr(OH)₂): 2000 mol/L
    • Volume of nitric acid solution: 350 mL
    • Concentration of nitric acid (HNO₃): 100 mol/L

Chemical Characteristics

  • Strontium Hydroxide (Sr(OH)₂):
    • Strong base.
    • Dissociates to release hydroxide ions (OH⁻).
    • Each mole of Sr(OH)₂ releases two moles of OH⁻:
    • Dissociation Equation:
      ext{Sr(OH)}_2
      ightarrow ext{Sr}^{2+} + 2 ext{OH}^{-}

Calculation Steps

Step 1: Calculate Initial Moles of OH⁻

  • Formula Used:
    extMolesofOH−=extConcentration<em>extSr(OH)</em>2imesextVolume<em>extSr(OH)</em>2imesextconversionfactorext{Moles of OH}^- = ext{Concentration}<em>{ ext{Sr(OH)}</em>2} imes ext{Volume}<em>{ ext{Sr(OH)}</em>2} imes ext{conversion factor}

  • Calculation:

    • Concentration of Sr(OH)₂ = 2000 mol/L
    • Volume of Sr(OH)₂ = 0.150 L

    - Conversion Factor: 2 moles of OH⁻ / 1 mole of Sr(OH)₂

    extMolesofOH−=2000extmolextLimes2extmolOH−extmolSr(OH)2imes0.150extL=0.6extmolesext{Moles of OH}^- = 2000 \frac{ ext{mol}}{ ext{L}} imes 2 \frac{ ext{mol OH}^-}{ ext{mol Sr(OH)}_2} imes 0.150 ext{ L} = 0.6 ext{ moles}

  • Conclusion:

    • Initial moles of OH⁻ added = 0.6 moles.
    • Result is expressed to three significant figures (0.6).
Step 2: Calculate Initial Moles of H⁺

  • Source of H⁺:

    • Comes from nitric acid
    • Each mole of HNO₃ releases one proton (H⁺).

  • Formula Used:
    extMolesofH+=extConcentration<em>extHNO</em>3imesextVolume<em>extHNO</em>3ext{Moles of H}^+ = ext{Concentration}<em>{ ext{HNO}</em>3} imes ext{Volume}<em>{ ext{HNO}</em>3}

  • Calculation:

    • Concentration of HNO₃ = 100 mol/L

    - Volume of HNO₃ = 0.350 L

    extMolesofH+=100extmolextLimes0.350extL=0.350extmolesext{Moles of H}^+ = 100 \frac{ ext{mol}}{ ext{L}} imes 0.350 ext{ L} = 0.350 ext{ moles}

  • Conclusion:

    • Initial moles of H⁺ added = 0.350 moles.
    • Result is expressed to three significant figures (0.350).
Step 3: Determine Limiting and Excess Reactant

  • Comparison:

    • Moles of OH⁻ = 0.6 moles
    • Moles of H⁺ = 0.350 moles

  • Conclusion:

    • Since 0.6 moles of OH⁻ > 0.350 moles of H⁺, OH⁻ is in excess.
    • Calculate excess moles of OH⁻:
    • extExcessOH−=0.6−0.350=0.250extmolesOH−ext{Excess OH}^- = 0.6 - 0.350 = 0.250 ext{ moles OH}^-
Step 4: Calculate Concentration of OH⁻

  • Total Volume of Mixture:

    • Volume of Sr(OH)₂ = 150 mL = 0.150 L
    • Volume of HNO₃ = 350 mL = 0.350 L
    • Total Volume = 0.150 L + 0.350 L = 0.500 L

  • Formula Used to Calculate Concentration:
    extConcentrationofOH−=extExcessmolesofOH−extTotalVolumeext{Concentration of OH}^- = \frac{ ext{Excess moles of OH}^-}{ ext{Total Volume}}

  • Calculation:

    extConcentrationofOH−=0.250extmoles0.500extL=0.500extmol/Lext{Concentration of OH}^- = \frac{0.250 ext{ moles}}{0.500 ext{ L}} = 0.500 ext{ mol/L}

Step 5: Calculate pOH

  • Formula Used:
    pOH=−extlog[extOH−]pOH = - ext{log}[ ext{OH}^-]

  • Calculation:

    pOH=−extlog(0.500)=0.301pOH = - ext{log}(0.500) = 0.301

  • Conclusion:

    • Result is expressed to three significant figures (0.301).
Step 6: Calculate pH

  • Formula Used:
    pH=14−pOHpH = 14 - pOH

  • Calculation:

    pH=14−0.301=13.699pH = 14 - 0.301 = 13.699

  • Conclusion:

    • pH of the final mixture = 13.699, which is expressed to three decimal places.
    • This indicates a highly basic solution due to the excess OH⁻ ions.

Final Result

  • The pH of the final mixture is 13.699, reflecting its basicity because of the excess hydroxide ions present after the reaction.