Question

"How would you make 100 mL of a carbonic acid buffer at 0.5 M and pH = 6.0 using 1.0 M NaHCO3 and either 1.0 M NaOH or 1.0 M HCl and water?"

Answer 1

To make a carbonic acid buffer at 0.5 M and pH 6.0, mix equal molar amounts of NaHCO3 and H2CO3. For 100 mL at 0.5 M, mix 25 mL of 1.0 M HCl with 50 mL of 1.0 M NaHCO3 and dilute to 100 mL with water.

Step-by-step explanation:

To make 100 mL of a carbonic acid buffer at 0.5 M and pH = 6.0 using 1.0 M NaHCO3 and either 1.0 M NaOH or 1.0 M HCl, you first need to understand the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

For carbonic acid (pKa ≈ 6.1), when pH is 6.0:

6.0 = 6.1 + log([NaHCO3]/[H2CO3])

This suggests the ratio of [NaHCO3] to [H2CO3] should be close to 1:1. Solving for the concentrations:

log([NaHCO3]/[H2CO3]) = -0.1

[NaHCO3]/[H2CO3] ≈ 0.79

If you want a 0.5 M buffer, you'll need close to 0.25 M NaHCO3 and 0.25 M H2CO3.

To get H2CO3, you can add HCl to NaHCO3 since H2CO3 is not stable:

NaHCO3 + HCl → H2CO3 + NaCl

For 100 mL at 0.25 M, you would need 25 mmol of HCl. You would take 25 mL of 1.0 M HCl (because 25 mL × 1.0 M = 25 mmol) and add it to 50 mL of 1.0 M NaHCO3 (which provides 50 mmol NaHCO3) to keep the ratio. Then dilute to 100 mL with water.

Answer 2

You first start by weighing a quatity of NaHCO3 which is by calculating the molecular mass of the salt and then multiply it with the molarity given which is 0.5 M, the gram/mol gotten is then dissolve in some water, add 0.1M NaOH dropwise until the pH is 9.8. Transfer quantitatively to a 100 mL volumetric flask and dilute to the mark. Mix thoroughtly.

Step-by-step explanation:

molar mass of Na HCO3= 84.01

molarity given= 0.5 M

to get the g/mol to dissolve in 1000 mL , 0.5 x 84.01 =42.005 g/mol=1 L

to get 100 mL , 42.002 divide by 10

=4.2005g/mol