Acid And Bases Test: Titration Problems In Chemistry

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  • 1/8 Questions

    The Concentration of HCl, a strong acid is 0.5 M. If 20 ml HCl is needed to titrate 40 mL of NaOH, what is the Concentration of NaOH?

    • 0.25 M NaOH
    • 0.5 M NaOH
    • 1 M NaOH
    • 0.1 M NaOH
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About This Quiz

Titration is a process of slowly adding one solution of a known concentration to a known volume of an unknown concentration until the reaction gets neutralized. This trivia quiz is based on the titration problem of acids and bases that we learned and had some practice in the lab this week. See how much you understood by taking this test!

Acid And Bases Test: Titration Problems In Chemistry - Quiz

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  • 2. 

    What is the purpose of an indicator in the solution with the unknown concentration?

    • It tells when there is enough acid in the solution.

    • It tells when the equivalence point is obtained.

    • It tells when there is enough base in the solution

    Correct Answer
    A. It tells when the equivalence point is obtained.
    Explanation
    An indicator is used in the solution with the unknown concentration to determine when the equivalence point is obtained. The equivalence point is the point in a chemical reaction where the stoichiometrically equivalent amounts of reactants have been mixed together. By using an indicator, which is a substance that undergoes a color change at a specific pH or in the presence of certain ions, the user can visually determine when the solution has reached the equivalence point. This allows for accurate titration and determination of the unknown concentration.

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  • 3. 

    A 15.5 ml sample of 0.215 M KOH was titrated with a weak acid. It took 21.2 mL of the acid to reach the equivalence point. What is the molarity of the acid?

    • 0.157 M acid

    • 0.29 M acid

    • 0.0157 M acid

    • 0.029 M acid

    Correct Answer
    A. 0.157 M acid
    Explanation
    The molarity of the acid can be determined using the concept of stoichiometry and the volume of the acid used in the titration. Since the volume and molarity of the base (KOH) are known, we can use the equation M1V1 = M2V2, where M1 is the molarity of the base, V1 is the volume of the base used, M2 is the molarity of the acid, and V2 is the volume of the acid used. Plugging in the known values, we have (0.215 M)(15.5 mL) = (M2)(21.2 mL). Solving for M2, we find that the molarity of the acid is 0.157 M.

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  • 4. 

    The Concentration of LiOH is 0.50 M. If 25 mL of LiOH is needed to titrate 40 mL of HNO3 (nitric acid) what is the Concentration of HNO3?

    • 0.03125 M HNO3

    • 0.8 M HNO3

    • 0.3125 M HNO3

    • 0.08 M HNO3

    Correct Answer
    A. 0.3125 M HNO3
    Explanation
    The concentration of HNO3 can be calculated using the formula:
    Concentration of HNO3 = (Volume of LiOH x Concentration of LiOH) / Volume of HNO3
    Substituting the given values:
    Concentration of HNO3 = (25 mL x 0.50 M) / 40 mL
    Simplifying:
    Concentration of HNO3 = 12.5 / 40
    Concentration of HNO3 = 0.3125 M HNO3

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  • 5. 

    If 20.0 mL of 0.100 M HCl is titrated with 19.5 mL of a NaOH solution. What is the molarity of the NaOH solution?

    • 0.0975 M NaOH

    • 0.0103 M NaOH

    • 0.103 M NaOH

    • 0.975 M NaOH

    Correct Answer
    A. 0.103 M NaOH
    Explanation
    The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this question, we are given the volume and molarity of the HCl solution, and we need to find the molarity of the NaOH solution. Since the reaction between HCl and NaOH is 1:1, the moles of HCl is equal to the moles of NaOH. We can calculate the moles of HCl by multiplying the volume (in liters) by the molarity. Then, we divide the moles of HCl by the volume (in liters) of NaOH solution to find the molarity of NaOH. The correct answer is 0.103 M NaOH.

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  • 6. 

    The Concentration of NaOH is 0.5 M if 20 ml is needed to titrate 35 mL of acid, what is the concentration of the acid?

    • 0.875 M of acid

    • 0.0029 M of the acid

    • 0.29 M of the acid

    • 0.00875 M of acid

    Correct Answer
    A. 0.29 M of the acid
    Explanation
    Since the name of the acid (and the equation) is not given, you must assume that it is a 1:1 ratio. Use equation: volume of acid x molarity of acid = volume of base x molarity of base (when the ratio is 1:1).

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  • 7. 

    Calculate the molarity of a HCl solution if 25.0 mL of the solution is  neutralized by 15.5 mL of 0.800 M NaOH:           

    • 0.248 M

    • 0.496 M

    • 1.29 M

    • 0.645 M

    Correct Answer
    A. 0.496 M
  • 8. 

    The Equation for the Neutralization reaction between solutions of Potassium hydroxide and Hydrochloric Acid is:

    • H+ (aq) + OH- (aq) --> H2O

    • KOH (aq) + HCl (aq) -->H2O (l) + KCl

    • POH (aq) + HClO (aq) --> H2O (l) + PClO (aq)

    • KOH (aq) + HCl (aq) -->K2Cl (aq)+ H2O (l)

    • KOH (aq) + HCl (aq) --> K2Cl (aq)+ H2O (l) + H+

    Correct Answer
    A. KOH (aq) + HCl (aq) -->H2O (l) + KCl
    Explanation
    The correct answer is KOH (aq) + HCl (aq) -->H2O (l) + KCl. This equation represents the neutralization reaction between potassium hydroxide (KOH) and hydrochloric acid (HCl), which results in the formation of water (H2O) and potassium chloride (KCl). In this reaction, the hydroxide ion (OH-) from KOH combines with the hydrogen ion (H+) from HCl to form water, while the potassium ion (K+) from KOH combines with the chloride ion (Cl-) from HCl to form potassium chloride.

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  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 12, 2009
    Quiz Created by
    Kwchiro
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