The part about dilution to 100.
For the NaOH/HCl part, it will be (H^+) = moles excess HCl/final volume in liters and pH from that.
For the buffer part (HAc/Ac^-), the dilution has no effect because with the Henderson-Hasselbalch equation you are diluting both the base and the acid th same amount and the ratio doesn't change.
pH = pKa + log (base)/(acid)
What is the pH of the solution created by combining 2.70 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
they ask you to find the pH for both the HCl and the HC2H3O2.
i understand the first question its just the second question that is confusing me
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