Question

You wish to prepare 100 mL (total volume) of a buffer solution that is 0.025 M in carbonic acid, pH 7.25. You have solid H2CO3 and solid NaHCO3. Calculate how many grams of H2CO3 and HCO3- you need to weigh out to prepare this solution.

This question has been posted before but can someone please explain where the 7.76, 8.76, and 1 came from?

7.25=6.36 +log (base/acid)
0.89= log (base/acid)
7.76=base/acid
base=(7.76/8.76) *0.025M/0.1L=0.2214M NaHCO3
acid=(1/8.76) * 0.025M/0.1L=0.0285M H2CO3
Need 0.1L*0.025M NaHCO3= 0.0025 moles NaHCO3
Convert that number of moles of NaHCO3 to grams and you should be good to go. (0.0025 * molar mass of NaHCO3)

Answer 1

Process:

• First: Find [base]/[Acid] relation.

pH = pKa + Log₁₀[base]/[Acid]

pH - pKa = Log₁₀[base]/[Acid]

10^{(pH - pKa)} = 10^{(Log₁₀[base]/[Acid])}

10^{(pH - pKa)} = [base]/[Acid]

[base]/[Acid] = 10^{(7.25 - 6.36)}

[base]/[Acid] = 7.76

• Second: Find Mass of acid value.

Volume of acid solution: 50 mL (0.05L)

Molar mass (H₂CO₃): 62 g.mol^-1

Concentration: 0.025M

Mass(H₂CO₃)=V_(L) x F_w x [Acid]

Mass(H₂CO₃)=0.05L x 62g.mol^-1 x 0.025 mol.L^-1

Mass(H₂CO₃)=0.0775g of solid H₂CO₃

• Third: Find Mass of base value.

Volume of base solution: 50 mL (0.05L)

Molar mass (NaHCO₃): 84 g.mol^-1

Concentration: ¿?

IF:

[base]/[Acid] = 7.76

THEN:

[base]= 7.76 x [Acid]

[base]= 7.76 x 0.025M

[base]= 0.194M

Mass(NaHCO₃)=V_(L) x F_w x [Base]

Mass(NaHCO₃)=0.05L x 84g.mol^-1 x 0.194 mol.L^-1

Mass(NaHCO₃)=0.8148g of solid NaHCO₃