Imagine that you are in chemistry lab and need to make 1.00 L of a solution with
ID: 995400 • Letter: I
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 7.00×102 M HCl, 100 mL of 5.00×102 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 82.0 mL of HCl and 90.0 mL of NaOH left in their original containers.
a) Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?
Explanation / Answer
First we write the reaction
H++OH=H2O
The pH is 3.11
The pH desired is 2.4
The [H] concentration you need is: 3.981x10-3 M
Initial moles=3.981x10-3+5x10-3=4.481x10-3 moles
moles=mol/L*L
Solve for L
L=(4.481x10-3 moles)/(7x10-2 M)=0.0640142 L= 64.014 mL
Therefore you need to add: 64.014 mL of HCl IN TOTAL.
you have already added 18 mL of HCl, so if you made the subtraction (64.014-18) = 46 mL. You need to add 46 mL.
H+ -OH Initial 1.26X10-3 React 5X10-4 Eq. 7.6X10-4