By how much will the pH change if 0.025 mol of HCl is added to 1.00 L of the buffer that contains 0.15 M HC 2 H 3 O 2 and 0.25 M C 2 H 3 O 2 – ? Use K a = 1.8 × 10 –5 for HC 2 H 3 O 2 .
Solution
Moles of HC2H3O2 = 0.15M*1L = 0.15moles
Moles of C2H3O2- = 0.25M*1L = 0.25moles
Moles of HCl added = 0.025moles
HCl will react with C2H3O2- to form HC2H3O2
HCl + C2H3O2- --> HC2H3O2 + Cl-
So, moles of C2H3O2- = 0.25moles-0.025moles = 0.225moles
Moles of HC2H3O2 = 0.15moles + 0.025moles = 0.175moles
pH = pKa + log[C2H3O2-]/[HC2H3O2]
Ka = 1.8*10 -5
pKa = -logKa = 4.74
pH = 4.74 + log0.225/0.175
= 4.85
.
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By how much will the pH change if 0-025 mol of HCl is added to 1-00 L.docx
1. By how much will the pH change if 0.025 mol of HCl is added to 1.00 L of the buffer that
contains 0.15 M HC 2 H 3 O 2 and 0.25 M C 2 H 3 O 2
–
? Use K a = 1.8 × 10 –5
for HC 2 H 3 O
2 .
Solution
Moles of HC2H3O2 = 0.15M*1L = 0.15moles
Moles of C2H3O2- = 0.25M*1L = 0.25moles
Moles of HCl added = 0.025moles
HCl will react with C2H3O2- to form HC2H3O2
HCl + C2H3O2- --> HC2H3O2 + Cl-
So, moles of C2H3O2- = 0.25moles-0.025moles = 0.225moles
Moles of HC2H3O2 = 0.15moles + 0.025moles = 0.175moles
pH = pKa + log[C2H3O2-]/[HC2H3O2]
Ka = 1.8*10 -5
pKa = -logKa = 4.74
pH = 4.74 + log0.225/0.175
= 4.85
![By how much will the pH change if 0.025 mol of HCl is added to 1.00 L of the buffer that
contains 0.15 M HC 2 H 3 O 2 and 0.25 M C 2 H 3 O 2
–
? Use K a = 1.8 × 10 –5
for HC 2 H 3 O
2 .
Solution
Moles of HC2H3O2 = 0.15M*1L = 0.15moles
Moles of C2H3O2- = 0.25M*1L = 0.25moles
Moles of HCl added = 0.025moles
HCl will react with C2H3O2- to form HC2H3O2
HCl + C2H3O2- --> HC2H3O2 + Cl-
So, moles of C2H3O2- = 0.25moles-0.025moles = 0.225moles
Moles of HC2H3O2 = 0.15moles + 0.025moles = 0.175moles
pH = pKa + log[C2H3O2-]/[HC2H3O2]
Ka = 1.8*10 -5
pKa = -logKa = 4.74
pH = 4.74 + log0.225/0.175
= 4.85](https://image.slidesharecdn.com/byhowmuchwillthephchangeif0-025molofhclisaddedto1-00l-230128090615-2433aadc/85/By-how-much-will-the-pH-change-if-0-025-mol-of-HCl-is-added-to-1-00-L-docx-1-320.jpg)
