Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.40.

You have in front of you

• 100 mL of 7.00×10⁻² 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 84.0 mL of HCl and 85.0 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.

Answer 1

Aim: To make IL solution with PH=2.40 We know, pH = - log lit]; cox) [H+] = 16 PH = 102:40 - 3.9exis Given solutions: 100 ml 7.0x102 M HCl Too ml 5.0x162 M Naolt. 84 ml Hel and 85 ml NaOH are lett Volume of Hol added = 100ml - 84 ml = 16ml volume of NaOH added = 100ml -85 ml = 15 ml. moles of Hel added = Modlarity & Volume in L. = 7.0x10² M x 16 ml. o ( 1 ) = 1+12 x103 moles. moles of Naoh added =5.0 x102mX 0.015L = 0.75 Neutralisation takes place moles of Hcl remained = (1012x103 – 0.75*103) moles = 0.37 x103 moles Total volume of the solution - IL molarity we require is 3.98x10²M le 3.98x103 moles in IL. Extra moles of Hd required = (3.98 x162 _ 0.37*10* moles = 3.61x162 moles moles = molarity < Volume (L) Volume = Moles = 361x10 moles < =0:515 7x10 L - Molarity 7.000 102 mol/ = 0.05157L. Volume = 0.05157 L x1000 ml = 51.57 ml ł 51.6 ml. Volume of Hol to be added = 51.6 ml.