Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.70. 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 90.0 mL of NaOH left in their original containers. Part A 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. View Available Hint(s)

Answer 1

We have to make 1 L solution of pH 2.70

pH = -log[H⁺] = 2,70

[H⁺] = 10^-2.70 M = 0.001995 M

Since the volume of solution is 1L, number of moles of H⁺ in 1L solution is 0.001995 mol

.

We have 100 mL 0.07 M HCl and 100 mL 0.05 M NaOH.

.

84.0 mL of HCl is left. Hence we have added 100-84.0 = 16.0 mL of HCl.

Number of moles of HCl in 16 mL = 0.07 X 16/1000 mol = 0.00112 mol

.

90.0 mL of NaOH is left. Hence we have added 100-90.0 = 10.0 mL of NaOH.

Number of moles of NaOH in 10 mL = 0.05 X 10/1000 mol = 0.0005 mol

.

0.0005 mol of NaOH has neutralized 0.0005 moles of HCl, hence the number of moles of HCl left are 0.00112 - 0.0005 mol = 0.00062 mol

Since HCl is strong acid, this 0.00062 mol HCl will completely dissociate yielding 0.00062 mol of H⁺ and 0.00062 mol of Cl^-.

We required 0.001995 mol of H⁺ and hence we need to add 0.001995-0.00062 mol = 0.001375 mol of H⁺

Since molarity of HCl is 0.07 M, 0.07 mol of HCl is contaoined in 1L=1000 mL of HCl solution

0.001375 mol HCl will b contained in 0.001375 X 1000/0.07 mL = 19.64 mL of HCl solution.

Therefore, we need to add 19.6 mL more HCl to the beaker (upto three significant figures)