Question

How many grams of NaHCO3 (FM 84.01 g/mol) should be mixed with Na2CO3 to produce a 1.00 L buffer solution with pH 9.50. The final concentration of Na2CO3 in this solution is 0.10 M. pKa1 = 6.37 and pKa2 = 10.33 for H2CO3.

Answer 1

Answer: 58.0 grams of sodium bicarbonate will be required.

Explanation: pH of buffer solutions is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

sodium carbonate acts as a base since it has carbonate ion where as sodium bicarbonate acts as an acid since it has bicarbonate ion that has a proton.

pKa2 value will be used here since we have sodium bicarbonate and not carbonic acid. Concentration of sodium carbonate is given as 0.10 M, pH is given as 9.50 and pKa2 is given as 10.33.

Let's plug in the values in Handerson equation and calculate the concentration of sodium bicarbonate.

$9.50=10.33+log(\frac{0.10}{x})$

(where $x$ is the concentration of sodium bicarbonate)

$9.50-10.33=log(\frac{0.10}{x})$

$-0.83=log(\frac{0.10}{x})$

taking antilog to both sides:

$10^-^0^.^8^3=\frac{0.10}{x}$

$0.145=\frac{0.10}{x}$

$x=\frac{0.10}{0.145}$

$x=0.69$

Concentration of sodium bicarbonate is 0.69 M. Volume of solution is given as 1.00 L. So, the moles of sodium bicarbonate will be 0.69 moles.

Molar mass of sodium bicarbonate is 84.0 gram per mol. Multiply the moles by molar mass to calculate the required grams of it.

$0.69mol(\frac{84.0g}{1mol})$

= 57.96 g

So, 57.96 g which is almost 58.0 g of $NaHCO_3$ will be required.

Answer 2

The required amount of NaHCO3 required to obtain a buffer solution with pH of 9.50 will be 56.80 g

pH relationship for a buffer solution :

$pH = pKa + log(\frac{base}{acid}$

• pH = 9.50
• pKa = 10.33
• Concentration of base = 0.10
• Concentration of acid = a

$9.50 = 10.33 + log(\frac{0.10}{a}$

$9.50 - 10.33 = log(\frac{0.10}{a}$

$-0.83 = log(\frac{0.10}{a}$

Take the inverse log of both sides

$10^{-0.83} = \frac{0.10}{a}$

$0.1479 = \frac{0.10}{a}$

$a = \frac{0.10}{0.1479}$

$a = 0.676$

• Concentration = mole × Molar mass
• Molar mass of NaHCO3 = 84.01 g/mol

Concentration of NaHCO3 = 84.01 × 0.676 = 56.80 g

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