3. Calculate the pH of a solution that contains 0.250 M HCOOH (formic acid) and 0.100 M NaCOOH (sodium formate). Given Ka = 1.8 x 10 - 4 for HCOOH. Please include the ICE table.
HCOOH Ka = 1.8 x 10 - 4
pKa = -log Ka = -log (1.8 x 10 - 4)
= 3.74
HCOOH + NaOH ------------------> NaCOOH + H2O
0.35 0.1 0 0 --------------- initial
-0.1 -0.1 +0.1 +0.1 ------------ change
0.25 0 0.1 0.1 ----------- equilibrium
pH = pKa + log [salt / acid]
= 3.74 + log [0.1 / 0.250]
= 3.34
pH = 3.34
3. Calculate the pH of a solution that contains 0.250 M HCOOH (formic acid) and 0.100...
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