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A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of?
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A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titr...
Calculation of (H') of the solution after the addition of NaOH:
To calculate the (H') of the solution after the addition of NaOH, we need to consider the stoichiometry of the reaction between the monoprotic acid and NaOH. The reaction can be represented as follows:
Monoprotic acid + NaOH → Salt + Water
Since the acid is monoprotic, it can donate only one proton (H+) to the base. Therefore, the stoichiometry of the reaction is 1:1.
Determination of moles of acid:
To determine the moles of acid present in the solution, we can use the formula:
Moles of acid = Concentration of acid × Volume of acid solution
Given that the concentration of the acid is 0.1 M and the volume of the acid solution is 50 mL (0.05 L), we can calculate the moles of acid as follows:
Moles of acid = 0.1 M × 0.05 L = 0.005 moles
Determination of moles of NaOH:
To determine the moles of NaOH added to the solution, we can use the formula:
Moles of NaOH = Concentration of NaOH × Volume of NaOH solution
Given that the concentration of NaOH is 0.1 M and the volume of NaOH solution is 50 mL (0.05 L), we can calculate the moles of NaOH as follows:
Moles of NaOH = 0.1 M × 0.05 L = 0.005 moles
Determination of moles of excess NaOH:
Since the stoichiometry of the reaction is 1:1, the moles of NaOH added are equal to the moles of acid present in the solution. Therefore, all the NaOH added will react with the acid, leaving no excess NaOH.
Determination of moles of salt formed:
Since the stoichiometry of the reaction is 1:1, the moles of salt formed will be equal to the moles of NaOH added. Therefore, the moles of salt formed can be calculated as follows:
Moles of salt = Moles of NaOH = 0.005 moles
Calculation of (H') of the solution:
The (H') of the solution can be calculated using the formula:
(H') = Moles of acid / Total volume of solution
The total volume of the solution is the sum of the volumes of the acid and NaOH solutions, which is 50 mL + 50 mL = 100 mL = 0.1 L.
Therefore, the (H') of the solution can be calculated as follows:
(H') = 0.005 moles / 0.1 L = 0.05 M
Thus, the (H') of the solution after the addition of NaOH is 0.05 M.
To calculate the (H') of the solution after the addition of NaOH, we need to consider the stoichiometry of the reaction between the monoprotic acid and NaOH. The reaction can be represented as follows:
Monoprotic acid + NaOH → Salt + Water
Since the acid is monoprotic, it can donate only one proton (H+) to the base. Therefore, the stoichiometry of the reaction is 1:1.
Determination of moles of acid:
To determine the moles of acid present in the solution, we can use the formula:
Moles of acid = Concentration of acid × Volume of acid solution
Given that the concentration of the acid is 0.1 M and the volume of the acid solution is 50 mL (0.05 L), we can calculate the moles of acid as follows:
Moles of acid = 0.1 M × 0.05 L = 0.005 moles
Determination of moles of NaOH:
To determine the moles of NaOH added to the solution, we can use the formula:
Moles of NaOH = Concentration of NaOH × Volume of NaOH solution
Given that the concentration of NaOH is 0.1 M and the volume of NaOH solution is 50 mL (0.05 L), we can calculate the moles of NaOH as follows:
Moles of NaOH = 0.1 M × 0.05 L = 0.005 moles
Determination of moles of excess NaOH:
Since the stoichiometry of the reaction is 1:1, the moles of NaOH added are equal to the moles of acid present in the solution. Therefore, all the NaOH added will react with the acid, leaving no excess NaOH.
Determination of moles of salt formed:
Since the stoichiometry of the reaction is 1:1, the moles of salt formed will be equal to the moles of NaOH added. Therefore, the moles of salt formed can be calculated as follows:
Moles of salt = Moles of NaOH = 0.005 moles
Calculation of (H') of the solution:
The (H') of the solution can be calculated using the formula:
(H') = Moles of acid / Total volume of solution
The total volume of the solution is the sum of the volumes of the acid and NaOH solutions, which is 50 mL + 50 mL = 100 mL = 0.1 L.
Therefore, the (H') of the solution can be calculated as follows:
(H') = 0.005 moles / 0.1 L = 0.05 M
Thus, the (H') of the solution after the addition of NaOH is 0.05 M.
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A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of?
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A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of?.
A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 mL solution of 0.1 M monoprotic acid (K,- 1 109 at 298 K) is titrated with 0.1 M NaOH. 50 mL of NaOH at this temperature. at 298 K. Calculate the (H') of the solution after the addition of?.
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