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 7.00×10−2 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 81.0 mL of HCl and 88.0 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?

Answer 1

Initial pH

Vol of HCl added = 100 - 81 = 19 ml

Vol of NaOH added = 100 - 88 = 12 ml

after neutralization,

concentration of HCl remaining = (19-12) ml x 7 x 10^-2 M/(81 + 88) = 2.90 x 10^-3 M

pH = -log[H+] = 2.54

Concentration present [H+] in 1L = 2.90 x 10^-3 M x 169 ml/1000 ml = 4.901 x 10^-4 M in 1L

Now For pH = 2.80

concentration needed [H+] = 1.58 x 10^-3 M in 1 L

Net balance HCl needed = 1.58 x 10^-3 - 4.901 x 10^-4 = 1.09 x 10^-3 M

Volume of HCl to be added = 1.09 x 10^-3 M x 1000 ml/7.0 x 10^-2 M = 15.57 ml

So we have to add another 15.57 ml of HCl to get a pH 2.80 solution