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120 mL of 0.1 M H2SO4 solution is mixed with W 180 mL of 0.1 M NaOH solution. The concentration of H ion in final solution is?
Verified Answer
120 mL of 0.1 M H2SO4 solution is mixed with W 180 mL of 0.1 M NaOH so...
Molarity = no of moles of solute[H+]/ volume of solution
No of moles of [H+] for HCl = 0.3*0.1 = 0.03
No of moles of [H+] for H2SO4 = 0.2*2*0.3 = 0.12 (since there are two moles of H+ in a mole H2SO4
Total no of moles of [H+] = 0.15
Total volume of solution = 300ml = 0.3L
Thus Molarity = 0.15/0.3 = 0.5moles/L
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Most Upvoted Answer
120 mL of 0.1 M H2SO4 solution is mixed with W 180 mL of 0.1 M NaOH so...
Calculation:

Given:
Volume of H2SO4 solution (V1) = 120 mL
Concentration of H2SO4 solution (C1) = 0.1 M

Volume of NaOH solution (V2) = 180 mL
Concentration of NaOH solution (C2) = 0.1 M

We need to find the concentration of H+ ion in the final solution.

Step 1: Determine the moles of H2SO4 and NaOH

Number of moles of H2SO4 (n1) = volume (V1) x concentration (C1)
= 0.120 L x 0.1 M
= 0.012 moles

Number of moles of NaOH (n2) = volume (V2) x concentration (C2)
= 0.180 L x 0.1 M
= 0.018 moles

Step 2: Identify the limiting reagent

To determine the limiting reagent, we compare the moles of H2SO4 and NaOH. The reactant with fewer moles will be the limiting reagent.

In this case, H2SO4 has 0.012 moles and NaOH has 0.018 moles. Therefore, H2SO4 is the limiting reagent.

Step 3: Determine the moles of H+ ion

H2SO4 dissociates into 2 H+ ions and 1 SO4^2- ion.
Therefore, the number of moles of H+ ions produced from H2SO4 = 2 x 0.012 = 0.024 moles

Step 4: Determine the volume of the final solution

The final volume of the solution is the sum of the volumes of H2SO4 and NaOH solutions.
Final volume (Vf) = V1 + V2
= 0.120 L + 0.180 L
= 0.300 L

Step 5: Calculate the concentration of H+ ion in the final solution

Concentration of H+ ion (Cf) = moles of H+ ion (0.024) / Final volume (0.300 L)
= 0.08 M

Therefore, the concentration of H+ ion in the final solution is 0.08 M.

Explanation:
- The given problem involves the mixing of H2SO4 and NaOH solutions.
- The number of moles of H2SO4 and NaOH is determined using the formula: moles = volume x concentration.
- The limiting reagent is identified by comparing the moles of H2SO4 and NaOH. In this case, H2SO4 is the limiting reagent as it has fewer moles.
- The number of moles of H+ ions produced from H2SO4 is calculated by considering its dissociation.
- The final volume of the solution is determined by adding the volumes of H2SO4 and NaOH solutions.
- Finally, the concentration of H+ ion in the final solution is calculated by dividing the moles of H+ ions by the final volume of the solution.
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120 mL of 0.1 M H2SO4 solution is mixed with W 180 mL of 0.1 M NaOH solution. The concentration of H ion in final solution is?
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120 mL of 0.1 M H2SO4 solution is mixed with W 180 mL of 0.1 M NaOH solution. The concentration of H ion in final solution is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 120 mL of 0.1 M H2SO4 solution is mixed with W 180 mL of 0.1 M NaOH solution. The concentration of H ion in final solution is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 120 mL of 0.1 M H2SO4 solution is mixed with W 180 mL of 0.1 M NaOH solution. The concentration of H ion in final solution is?.
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