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 7.00×10−2mol L−1 HCl, 100 mL of 5.00×10−2mol L−1 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 89.0 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

Sol . As pH = - log[H+]   

so , [H+] = antilog(-pH) = antilog(-2.80) = 1.584 × 10-3 M

and , Volume of solution = 1L

So , moles of HCl  needed = Molarity × volume = 1.584 × 10-3 ×  1 = 1.584 × 10-3 mol

Now , volume of HCl used in the lab = 100 - 81 = 19 mL = 0.019 L

Molarity of HCl = 7.00 × 10-2 M

So , moles of HCl used in the lab = molarity × volume

= 7.00 × 10-2 × 0.019 = 1.33 × 10-3 mol

and , Volume of NaOH used in the lab = 100 - 89 = 11 mL = 0.011 L

Molarity of NaOH = 5.00 × 10-2 M

So , moles of NaOH used in the lab = molarity × volume = 5.00 × 10-2 × 0.011 = 0.55 × 10-3 mol

As Reaction : HCl + NaOH ---> NaCl + H2O

So , 1 mole of HCl combines with 1 mole of NaOH .

and , moles of HCl present after reaction = moles of HCl used in the lab - moles of NaOH used in the lab = 1.33 × 10-3 - 0.55 × 10-3

= 0.78 × 10-3 mol

Therefore ,

moles of HCl extra required = moles of HCl needed - moles of HCl present after reaction = 1.584 × 10-3 - 0.78 × 10-3 = 0.804 × 10-3 mol

As molatity of HCl = 7.00 × 10-2 M

So , Volume of HCl extra required = moles / molarity = (0.804 × 10-3) / (7.00 × 10-2) = 0.0115 L

= 0.0115   × 1000 = 11.5  mL

Therefore , answer is 11.5 mL