A 283 g silver figure of a polar bear is dropped into the 217 g aluminum cup of a well insulated calorimeter Containing 263 g of
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The question is incomplete, here is the complete question:
Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴
a) 2.099
b) 10.463
c) 3.546
d) 2.307
e) 3.952
<u>Answer:</u> The pH of the solution is 3.546
<u>Explanation:</u>
We are given:
Moles of formic acid = 0.370 moles
Moles of sodium formate = 0.230 moles
Volume of solution = 1 L
To calculate the molarity of solution, we use the equation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
= negative logarithm of acid dissociation constant of formic acid = 3.75
pH = ?
Putting values in above equation, we get:
Hence, the pH of the solution is 3.546