You have in front of you 100 mL of 6.00×10^ 2 M HCl, 100 mL of 5.00×10^ 2 M NaOH
ID: 1034510 • Letter: Y
Question
You have in front of you
100 mL of 6.00×10^2M HCl,
100 mL of 5.00×10^2M NaOH, and
plenty of distilled water.
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.
Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.50.You have in front of you
100 mL of 6.00×10^2M HCl,
100 mL of 5.00×10^2M 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 80.0 mL of HCland 90.0 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.
Explanation / Answer
answer- Okay, so we start by finding the number of moles of H+ already present in the solution.
Do this by multiplying the molarity of the acid by the volume added, and then subtract the number of moles of -OH added (molarity of the base times the volume added). Thus, we get:
(0.06M HCl)(0.020L)-(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+ concentration from the pH:
[H+] = 10^-2.50
= 0.003162...M
Since this is the number of moles per liter, and we have 1 liter, we know that the total number of moles of H+ will be 0.003162... mol. By subtracting the number of moles already in the solution from the number that need to be in the solution at the end, we get the amount that needs to be added:
0.003162mol - 6.4x10^-4 mol = 0.0025222 mol H+ needed.
Thus, (0.06M HCl)(? L)= 0.0025222mol H+
Solving for L, we get 0.042037L
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