What is the pH of the solution ?

Question

created by combining 12.50 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?

Answer 1

in all of these cases you have an excess of NaOH, it will determine the pH:

find moles:

0.0125 litres @ 0.10mol/litre = 0.00125 moles NaOH

0.008 litres @ 01. mol/litre = 0.0008 moles acid  (HCl or Acetic)

the difference shows that there is 0.00045 moles of NaOH left

your answers:

(with 8.00 mL of the 0.10 M HCl(aq), a total of 20.5 ml's):

0.00045 moles OH / 0.0205 litres= 0.02195 molar OH

0.02195 molar OH = = => pOH = 1.66

@ pH = 14 -pOH

first answer;  pH = 12.34

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with 8.00 mL of the 0.10 M Acetic(aq), a total of 20.5 ml's):

0.00045 moles OH / 0.0205 litres= 0.02195 molar OH

0.02195 molar OH = = => pOH = 1.66

@ pH = 14 -pOH

first answer;  pH = 12.34

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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

for both HCl & Acetic acids now in 112.50 ml:

0.00045 moles OH / 0.1125 litres= 0.004 molar OH

0.004 molar OH = = => pOH = 2.40

@ pH = 14 -pOH

first answer;  pH = 11.60