Part A Imagine that you are in chemistry lab and need to make 1.00 L of a soluti
ID: 704344 • Letter: P
Question
Part A Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.40 Assuming the final solution will be diluted to 1.00 L, how much more HCI should you add to achieve the desired pH? Express your answer to throe significant figures and include the appropriate units You have in front of you 100 mL of 7.00x102 MHCL, View Available Hint(s) 100 mL of 5.00*102 MNaOH, 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 the wrong Value Units cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 83.0 mL of HCI and 90.0 mL of NaOH left in theirExplanation / Answer
Answer
47.0ml
Explanation
pH = - log[H+]
-log[H+] = 2.40
[H+] = 3.98×10-3M
No of moles of HCl required = 3.98×10-3M
Concentration of given HCl solution = 0.07M
Volume of HCl solution required to get 0.00398M of HCl =(1000ml/0.0700mol)×0.00398mol = 56.86ml
moles of NaOH added = (0.05mol/1000ml)×10ml = 0.0005mol
No of moles HCl required to neutalize 0.0005 moles of NaOH = 0.0005mol
Volume of 0.07M HCl solution containing 0.0005mol of HCl = (1000ml/0.07mol)×0.0005mol= 7.14ml
Therefore
Total Volume 0.07M HCl required = 56.86ml+ 7.14ml = 64ml
Volume of 0.07M HCl solution added = 100ml - 83ml = 17ml
Volume of 0.07M HCl further to be added = 64ml - 17ml = 47ml