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kumpel [21]
4 years ago
15

How many grams of sodium formate, NaCHO2, would have to be dissolved in 3.0 L of 0.12 M formic acid (pKa 3.74) to make the solut

ion a buffer for pH 5.30?
Chemistry
1 answer:
kolbaska11 [484]4 years ago
5 0

Answer:

Mass_{sodium\ formate}= 889.57\ g

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the acidic buffer solution as:

pH=pK_a+log\frac{[salt]}{[acid]}

Given that:-

[Acid] = 0.12 M

Volume = 3.0 L

pKa = 3.74

pH = 5.30

So,  

5.30=3.74+log\frac{[sodium\ formate]}{0.12}

Solving, we get that:-

[Sodium formate] = 4.36 M

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,

Moles =Molarity \times {Volume\ of\ the\ solution}

So, Moles of sodium formate = 4.36*3.0 moles = 13.08 moles

Molar mass of sodium formate = 68.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

13.08\ mole= \frac{Mass}{68.01\ g/mol}

Mass_{sodium\ formate}= 889.57\ g

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