Question

14 grams of impure koh ( potassium hydroxide ) was dissolved in water to make 100...

14 grams of impure koh ( potassium hydroxide ) was dissolved in water to make 100 ml of solution. the solution was then titrated with 1.26 m nitric acid solution. 40.0 ml of the acid required 25.0 ml of alkali solution. determine the percentage koh of the sample

0 0
Add a comment Improve this question Transcribed image text
Request Professional Answer

Request Answer!

We need at least 10 more requests to produce the answer.

0 / 10 have requested this problem solution

The more requests, the faster the answer.

Request! (Login Required)


All students who have requested the answer will be notified once they are available.
Know the answer?
Add Answer to:
14 grams of impure koh ( potassium hydroxide ) was dissolved in water to make 100...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions
  • Part A A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a...

    Part A A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.7 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l) Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following...

  • QUESTION 1: A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a...

    QUESTION 1: A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 12.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l) QUESTION 2: Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following...

  • A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution...

    A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 15.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)?K2SO4(aq)+2H2O(l) part b: Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents...

  • Part A A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a...

    Part A A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 19.7 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l) Express your answer with the appropriate units. Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a...

  • A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of...

    A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the resulting solution is titrated with 2.50 M solution of HCl. The indicator changes colour, signalling that the equivalence point has been reached after 17.0 mL of the hydrochloric acid solution has been added. Determine the molality of this alkali metal hydroxide solution and What is the mass percentage of the metal hydroxide in the solution? Determine...

  • A 1.20 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water...

    A 1.20 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water and titrated with a a 0.485 M aqucous potassium hydroxide solution. It is observed that after 7.33 milliliters of potassium hydroxide have been added, the pH is 4.515 and that an additional 13.3 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? (2) What is the value of K, for the...

  • A 0.843 gram sample of an unknown monoprotic acid is dissolved in 40.0 mL of water...

    A 0.843 gram sample of an unknown monoprotic acid is dissolved in 40.0 mL of water and titrated with a a 0.496 M aqueous potassium hydroxidesolution. It is observed that after 5.16 milliliters of potassium hydroxide have been added, the pH is 9.092 and that an additional 9.84 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? _______g/mol (2) What is the value of Ka for the...

  • A 0.843 gram sample of an unknown monoprotic acid is dissolved in 40.0 mL of water and titrated w...

    A 0.843 gram sample of an unknown monoprotic acid is dissolved in 40.0 mL of water and titrated with a a 0.496 M aqueous potassium hydroxidesolution. It is observed that after 5.16 milliliters of potassium hydroxide have been added, the pH is 9.092 and that an additional 9.84 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? _______g/mol (2) What is the value of Ka for the...

  • Could someone answer the following questions? 1) 11.2 g of potassium hydroxide (KOH) is dissolved in...

    Could someone answer the following questions? 1) 11.2 g of potassium hydroxide (KOH) is dissolved in sufficient water to make 1 L of solution. What is the concentration of KOH in the solution? a. 0.02 mol/L b. 0.1 mol/L c. 0.2 mol/L d. 0.01 mol/L 2) If the solution is neutral, which of the following must be true? a. [OH-] = [H2O] b. [H+] > [OH-] c. [H+] < [OH-] d. [H+] = [OH-] 3)A student titrates 0.5222 grams of...

  • Potassium hydroxide is very soluble in water, resulting in extremely basic solutions. A 121 g sample...

    Potassium hydroxide is very soluble in water, resulting in extremely basic solutions. A 121 g sample KOH is dissolved in water at 25 °C to make up 100.0 mL of solution. The molar mass of KOH is 56.11 3. What is the pH of the solution at 25.0°C? • Round the answer to two decimal places.

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT