Question
In: Chemistry
Calculate the pH of each of the solutions and the change in pH to 0.01 pH...
Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.78-M HCl to 440. mL of each of the following solutions.
Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.
a) water
| pH before mixing = | |
| pH after mixing= | |
| pH change = |
b) 0.154 M C2H3O21-
| pH before mixing = | |
| pH after mixing= | |
| pH change = |
c) 0.154 M HC2H3O2
| pH before mixing = | |
| pH after mixing= | |
| pH change = |
d) a buffer solution that is 0.154 M in each C2H3O21- and HC2H3O2
| pH before mixing = | |
| pH after mixing= | |
| pH change = |
Solutions
Expert Solution
a)
pH before mixing = 7.00
resultant HCl molarity = 3.78 x 10 / (440 +10) = 0.084 M
pH = -log [H+]
pH = 1.08
pH after mixing = 1.08
change in pH = -5.92
b)
it is the salt of weak acid and strong base . so pH >7
pH = 7 + 1/2 [pKa + log C]
pH = 7 + 1/2 [4.74 + log 0.154]
pH = 8.96
pH before mxing = pH = 8.96
millimoles of CH3COO- = 440 x 0.154 = 67.76
millimoles of HCl = 3.78 x 10 = 37.8
CH3COO- + HCl -----------------> CH3COOH
67.76 37.8 0
29.96 0 37.8
pH = pKa + log [CH3COO- / CH3COOH]
pH = 4.74 + log (37.8 / 29.96)
pH = 4.64
pH after mixing = 4.64
change in pH = - 4.32
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