Acid Base Titration Problems, Basic Introduction, Calculations, Examples, Solution Stoichiometry

The Organic Chemistry Tutor・2 minutes read

Two methods, equation (m1v1 = m2v2) and dimensional analysis, are used to solve acid-base titration problems, with specific ratios and concentrations calculated for different scenarios involving various acids and bases. The concentration of the solutions, along with volume requirements to neutralize them, are determined through the application of stoichiometry and double replacement reactions.

Insights

  • The acid-base titration problems can be solved using the equation m1v1 = m2v2 or dimensional analysis, with each method corresponding to the acid (m1v1) and base (m2v2) in the reaction, respectively.
  • The molar ratios between different acids and bases, such as HCl and NaOH (1:1) or barium hydroxide and HNO3 (1:2), impact the concentration calculations, with specific volumes required for neutralization based on these ratios.

Get key ideas from YouTube videos. It’s free

Recent questions

  • How are acid-base titration problems solved?

    Acid-base titration problems can be solved using two methods: an equation (m1v1 = m2v2) and dimensional analysis. The equation method involves identifying the acid (m1v1) and base (m2v2) components, with the concentration of the original solution calculated. The second method utilizes balanced chemical equations to determine the molar ratio between reactants, leading to concentration results.

  • What is the molar ratio between HCl and NaOH?

    The molar ratio between HCl and NaOH in a double replacement reaction is one-to-one, resulting in the same concentration of 0.362 moles per liter for both substances.

  • How is the concentration of barium hydroxide calculated?

    The concentration of barium hydroxide is calculated by considering the molar ratio between barium hydroxide and HNO3, which is one-to-two due to the two hydroxide ions in barium hydroxide. This calculation results in a concentration of 0.15 moles per liter for barium hydroxide.

  • What is the volume requirement to neutralize HNO3 with barium hydroxide?

    The volume requirement to neutralize HNO3 with barium hydroxide is determined to be 43.5 milliliters, calculated based on the concentration of barium hydroxide and the molar ratio between the two substances.

  • How is the concentration of a KOH solution determined?

    The concentration of a KOH solution is determined by titrating KHP with the KOH solution and calculating the mass of KHP neutralized. By using stoichiometry and the volume of KOH solution used, the concentration of the KOH solution is found to be 0.01 moles per liter.

Related videos

Summary

00:00

Acid-Base Titration Methods and Calculations

  • To solve acid-base titration problems, two methods can be used: an equation (m1v1 = m2v2) and dimensional analysis.
  • The acid in the problem is monoprotic, while the base has one hydroxide ion per formula unit.
  • The formula m1v1 corresponds to the acid, and m2v2 corresponds to the base in the equation m1v1 = m2v2.
  • The concentration of the original HCl solution is found to be 0.362 moles per liter using the equation method.
  • The balanced chemical equation for HCl and NaOH in a double replacement reaction is utilized in the second method.
  • The molar ratio between HCl and NaOH is one-to-one, leading to the same concentration result of 0.362 moles per liter.
  • For a problem involving barium hydroxide and HNO3, the molar ratio is one-to-two due to the two hydroxide ions in barium hydroxide.
  • The concentration of barium hydroxide is calculated to be 0.15 moles per liter, resulting in a volume requirement of 43.5 milliliters to neutralize the HNO3 solution.
  • In a scenario with KHP and NaOH, stoichiometry is used to find the mass of KHP neutralized by 32.57 milliliters of a 0.175 molar NaOH solution, resulting in 1.164 grams of KHP.
  • The concentration of KOH solution is determined to be 0.01 moles per liter after titrating 0.137 grams of KHP with 42.6 milliliters of KOH.
Channel avatarChannel avatarChannel avatarChannel avatarChannel avatar

Try it yourself — It’s free.