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If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?
- a)0.020 M
- b)0.100 M
- c)0.025 M
- d)0.050 M
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025...
M1V1 = M2V2
(0.025 M) (0.050 L) = (M2) (0.025 L)
M2 = 0.05 M
but, there are 2 H’s per H2SO4 so [H2SO4] = 0.025 M
(0.025 M) (0.050 L) = (M2) (0.025 L)
M2 = 0.05 M
but, there are 2 H’s per H2SO4 so [H2SO4] = 0.025 M
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Community Answer
If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025...
To determine the molarity of the sulfuric acid solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
In this reaction, one mole of sulfuric acid reacts with two moles of sodium hydroxide. The balanced equation tells us that the ratio of moles of sulfuric acid to moles of sodium hydroxide is 1:2.
Now, let's calculate the number of moles of sodium hydroxide used in the titration:
Moles of NaOH = volume (in L) x molarity
= 0.050 L x 0.025 mol/L
= 0.00125 mol
Since the ratio of moles of sulfuric acid to moles of sodium hydroxide is 1:2, the number of moles of sulfuric acid used in the titration is half that of sodium hydroxide:
Moles of H2SO4 = 0.00125 mol / 2
= 0.000625 mol
Next, we need to calculate the volume of the sulfuric acid solution used in the titration. The volume of the sulfuric acid solution is given as 25.0 mL, which is equivalent to 0.025 L.
Now, we can calculate the molarity of the sulfuric acid solution using the equation:
Molarity = moles / volume (in L)
= 0.000625 mol / 0.025 L
= 0.025 M
Therefore, the molarity of the sulfuric acid solution is 0.025 M. Thus, option C is the correct answer.
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
In this reaction, one mole of sulfuric acid reacts with two moles of sodium hydroxide. The balanced equation tells us that the ratio of moles of sulfuric acid to moles of sodium hydroxide is 1:2.
Now, let's calculate the number of moles of sodium hydroxide used in the titration:
Moles of NaOH = volume (in L) x molarity
= 0.050 L x 0.025 mol/L
= 0.00125 mol
Since the ratio of moles of sulfuric acid to moles of sodium hydroxide is 1:2, the number of moles of sulfuric acid used in the titration is half that of sodium hydroxide:
Moles of H2SO4 = 0.00125 mol / 2
= 0.000625 mol
Next, we need to calculate the volume of the sulfuric acid solution used in the titration. The volume of the sulfuric acid solution is given as 25.0 mL, which is equivalent to 0.025 L.
Now, we can calculate the molarity of the sulfuric acid solution using the equation:
Molarity = moles / volume (in L)
= 0.000625 mol / 0.025 L
= 0.025 M
Therefore, the molarity of the sulfuric acid solution is 0.025 M. Thus, option C is the correct answer.
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If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer?
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If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer?.
If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer?.
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If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?a)0.020 Mb)0.100 Mc)0.025 Md)0.050 MCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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