Imagine that you are in chemistry lab and need to make 1.00 LLof a solution with a pHpH of 2.60.
You have in front of you
- 100 mLmL of 6.00×10−2mol L−16.00×10−2mol L−1 HClHCl,
- 100 mLmL of 5.00×10−2mol L−15.00×10−2mol L−1 NaOHNaOH, and
- plenty of distilled water.
You start to add HClHCl to a beaker of water when someone asksyou a question. When you return to your dilution, you accidentallygrab the wrong cylinder and add some NaOHNaOH. Once you realizeyour error, you assess the situation. You have 83.0 mLmL of HClHCland 90.0 mLmL of NaOHNaOH left in their original containers.
Assuming the final solution will be diluted to 1.00 LL , howmuch more HClHCl should you add to achieve the desired pH?
