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Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2. You have in front of you 100 mL of 6.00102 M HCl, 100 mL...

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.60. You have in front of you 100 mL of 6.00×10−2 M HCl, 100 mL of 5.00×10−2 M NaOH, and plenty of distilled water. You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 81.0 mL of HCl and 85.0 mL of NaOH left in their original containers. Part A Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH? Express your answer to three significant figures and include the appropriate units.

You have 81.0 mL of HCl and 85.0 mL of NaOH left in their original containers. SO you have added 19 mL of HCl and 15 mL of... View the full answer

• See the right answer the volumes have been considered correctly
• rghosh1976
• Jul 20, 2016 at 5:02am
• In the other calculation the initial volumes are wrong
• rghosh1976
• Jul 20, 2016 at 5:03am
• Pl choose and rate
• rghosh1976
• Jul 20, 2016 at 5:03am
• HMM OK
• jack897
• Jul 20, 2016 at 5:06am

(0.06M HCl)(0.019L)-(0.05M NaOH)(0.010L)=6.4x10^-4 moles of unreacted H+ Next, we can calculate how much H+ should be in the... View the full answer

Great question... View the full answer

(0.06M HCl)(0.019L)-(0.05M NaOH)(0.010L)=6.4x10^-4 moles of unreacted H+
Next, we can calculate how much H+ should be in the solution once it's diluted to 1 L. This is
done by calculating the H+...

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