Imagine that you are in chemistry lab and need to make 1.00 L of a solution with
ID: 521500 • Letter: I
Question
Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.80. You have in front of you 100 mL of 7.00 times 10^-2 M HCl, 100 mL of 5.00 times 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 85.0 mL of HCl and 90.0 mL of NaOH left in their original containers. 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
Volume of HCl added = 100 - 85.0 = 15.0 mL
Volume of NaOH added = 100 - 90.0 = 10.0 mL
Molarity of HCl = 7.00 * 10-2 M
Molarity of NaOH = 5.00 * 10-2 M
The neutrlisation reaction between HCl and NaOH can be written as,
NaOH (aq.) + HCl (aq.) -------------> NaCl (aq.) + H2O (l)
since, MAVA > MBVB
Molariy of acid in the resulting solution = (MAVA - MBVB) / (VA + VB)
M = (0.0700 * 15.0 - 0.0500 * 10.0) / (15.0 + 10.0)
M = 0.0220 M
Therefore, using neutralisation formula,
M1 V1 = M2 V2
0.0220 * 25.0 = M2 * 1000.
M2 = 5.50 * 10-4 M