Imagine that you are in chemistry lab and need to make 1.00 L of a solution with
ID: 304602 • Letter: I
Question
Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.60.
You have in front of you
100 mL of 7.00×10?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 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
Since we have 80 mL HCl and 90 mL Naoh left in the beakers. this means we have poured 20 mL of HCl and 10 ml of NaOHinside the final solution.
Now, Calculating the pH of this solution will give us idea about how much HCl is required for pH = 2.60
20 mL of 7 * 10 -2 M HCl has 20 * 7 *10-2 mmoles of HCl= 1.4 mmoles
10 mL of 5 * 10-2 M NaOH has 5 * 10 * 10-2 mmoles of NaOH = 0.5 mmoles
0.5 mmoles of NaOH will be neutralised by 0.5 mmoles of HCl and (1.4 - 0.5 =0.9 mmoles ) will be left behind which corresponds to concentration of 0.9 mmoles /1000 ml (Volume of final solution) =0.0009 M
Coresponding pH =-LOG [H+] = -LOG [0.0009] = 3.0457
Now adding 22.75 mL of HCl , so total volume of HCL = 20 + 22.75 mL Of HCl = 42.75 mL
42.75 mL of 7 * 10 -2 M HCl has 42.75 * 7 * 10-2 m moles of HCl = 2.9925 mmoles
0.5 mmoles of NaOH will be neutralised by 0.5 mmoles of HCl and (2.9925 - 0.5 = 2.4925 mmoles ) will be left behind which corresponds to concentration of 2.4925 mmoles /1000 mL (Volume of final solution) =0.0024925 M
Coresponding pH =-LOG [H+] = -LOG [0.0024925] = 2.6033 approximately 2.60
Rounding off to three significant digits 22.75 becomes 22.8 mL of HCl