# Acid And Bases Test: Titration Problems In Chemistry

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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!

• 1.

### 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?

• A.

0.875 M of acid

• B.

0.0029 M of the acid

• C.

0.29 M of the acid

• D.

0.00875 M of acid

C. 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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• 2.

### 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?

• A.

0.157 M acid

• B.

0.29 M acid

• C.

0.0157 M acid

• D.

0.029 M acid

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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• 3.

### 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?

• A.

0.0975 M NaOH

• B.

0.0103 M NaOH

• C.

0.103 M NaOH

• D.

0.975 M NaOH

C. 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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• 4.

### 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?

• A.

0.25 M NaOH

• B.

0.5 M NaOH

• C.

1 M NaOH

• D.

0.1 M NaOH

A. 0.25 M NaOH
Explanation
The concentration of HCl is given as 0.5 M. The volume of HCl used in the titration is 20 mL, and it is used to titrate 40 mL of NaOH. Since HCl and NaOH react in a 1:1 ratio according to the balanced chemical equation, the moles of HCl used in the titration are equal to the moles of NaOH present in the solution. Therefore, the concentration of NaOH can be calculated by dividing the moles of NaOH by the volume of NaOH used in the titration. Since the volume of NaOH used in the titration is 40 mL, and the concentration of NaOH is equal to the concentration of HCl, which is 0.5 M, the concentration of NaOH is 0.25 M.

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• 5.

### 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?

• A.

0.03125 M HNO3

• B.

0.8 M HNO3

• C.

0.3125 M HNO3

• D.

0.08 M HNO3

C. 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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• 6.

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

• A.

It tells when there is enough acid in the solution.

• B.

It tells when the equivalence point is obtained.

• C.

It tells when there is enough base in the solution

B. 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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• 7.

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

• A.

H+ (aq) + OH- (aq) --> H2O

• B.

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

• C.

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

• D.

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

• E.

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

B. 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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• 8.

### 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:

• A.

0.248 M

• B.

0.496 M

• C.

1.29 M

• D.

0.645 M

B. 0.496 M

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• Current Version
• Mar 22, 2023
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• Apr 12, 2009
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