You wish to prepare 100 mL (total volume) of a buffer solution that is 0.025 M i

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)

Explanation / Answer

Process:

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

Volume of acid solution: 50 mL (0.05L)

Molar mass (H2CO3): 62 g.mol-1

Concentration: 0.025M

Mass(H2CO3)=V(L) x Fw x [Acid]

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

Mass(H2CO3)=0.0775g of solid H2CO3

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 Fw x [Base]

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

Mass(NaHCO3)=0.8148g of solid NaHCO3

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