Question

How many grams of Na2CO3 (FM 105.99) should be mixed with 4.61 g of NaHCO3 (FM 84.01) to produce a 100 mL of buffer with pH of 10.44? (The Ka's of carbonic acid are Ka1 = 4.46 ✕ 10−7 and Ka2 = 4.69 ✕ 10−11.)

Answer #1

**we know that**

**moles = mass / molar mass**

**so**

**moles of NaHC03 = 4.61 / 84.01**

**moles of NaHC03 = 0.0548744**

**we know that**

**for a buffer solution**

**pH = pKa + log [salt / acid ]**

**in this case**

**pH = pKa2 + log [ Na2C03 / NaHC03]**

**now**

**we know that**

**pKa2 = -log Ka2**

**so**

**pH = -log Ka2 + log [ Na2C03 / NaHC03]**

**10.44 = -log 4.69 x 10-11 + log [ Na2C03 /
NaHC03]**

**[ Na2C03 / NaHC03] = 1.2917**

**now**

**we know that**

**concentration = moles / volume**

**as the final volume is same**

**ratio of concentrations = ratio of moles**

**so**

**moles of Na2C03 / moles of NaHC03 = 1.2917**

**so**

**moles of Na2C03 / 0.0548744 = 1.2917**

**moles of Na2C03 = 0.070883**

**now**

**mass = moles x molar mass**

**so**

**mass of Na2C03 = 0.070883 x 105.99**

**mass of Na2C03 = 7.513**

**so**

**7.513 grams of Na2C03 is required**

How many grams of Na2CO3 (FM 105.99) should be mixed with 3.97 g
of NaHCO3 (FM 84.01) to produce a 100 mL of buffer with pH of
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