how many moles of sodium bicarbonate are contained in 1mL of a saturated solution of NaHCO3?

Answer 1

As for most solutes, the solubility of NaHCO3 is a function of temperature. Because you didn't specify a temperature, I'll assume you mean "room temperature" (25 C).

The solubility of NaHCO3 also depends on the partial pressure of CO2 that the solution is in equilibrium with. I'll assume you want to know the saturation concentration for a solution in equilibrium with normal air. Currently, the Earth's atmosphere has a partial pressure of CO2 equal to 3.8*10^-4. An interesting calculation (which I'll do below) is to see what the solubility of NaHCO3 would have been around the beginning of the 20th century, when the partial pressure of CO2 was more like 3.0 * 10^-4 (i.e., before the large addition of anthropogenic CO2 to the atmosphere from fossil fuel burning in the 20th century).

There are 5 chemical reactions involved in the solution equilibria of NaHCO3, and an equilibrium constant expression can be written for each reaction (see sources for the actual values of the equilibrium constants):

(1): NaHCO3(s) <-> Na+ + HCO3- K0 = [Na+][HCO3-] = 10^-0.4025
(2): CO2(g) + H2O(l) <-> H2CO3(aq) KH = [H2CO3]/pCO2 = 10^-1.474
(3): H2CO3(aq) <-> H+ + CO3- K1 = [H+][HCO3-]/[H2CO3] = 10^-6.37
(4): HCO3- <-> H+ + CO3-- K2 = [H+][CO3--]/[HCO3-] = 10^-10.25
(5): H2O <-> H+ + OH- Kw = [H+][OH-] = 10^-14

where the quantities in square brackets are the concentrations (strictly speaking, the thermodynamic activities) of the species, and pCO2 is the partial pressure of CO2.

We also have an equation expressing the fact that the solution must be electrically neutral (i.e., the positive and negative charges on the ions in solution must balance):

(6): [Na+] + [H+] = 2*[CO3--] + [HCO3-] + [OH-]

We have 7 unknowns (the partial pressure of CO2, and the concentrations of H+, OH-, Na+, H2CO3, HCO3-, CO3--) and 6 equations. If we specify the value of any one of the unknowns, we can solve for the other 6. The only unknown that it make sense to specify is pCO2.

We can rewrite and combine the above equibibrium constant expressions to eliminate all the unknowns in the charge-balance equation except [Na+] and pCO2.

[HCO3-] = K0.[Na+] (from (1))
[H2CO3] = pCO2*KH (from (2))
[H+] = K1*KH*pCO2*[Na+]/K0 (from (3) and (2))
[OH-] = Kw*K0/(K1*KH*pCO2*[Na+]) (from (5) and (2) and (3))
[CO3--] = K0^2 * K2/(K1*KH*pCO2*[Na+]^2) (from (4) and (3) and (2))

Plugging all these into the charge-balance equation (6) gives:

[Na+] + K1*KH*pCO2*[Na+]/K0 = 2*K0^2*K2/(K1*KH*pCO2*[Na+]^2) + K0/[Na+] + Kw*K0/(K1*KH*pCO2*[Na+])

Multiply through by pCO2*[Na+}^2 to clear the denominator of these unknowns:

PCO2*[Na+]^3 + K1*KH*(pCO2)^2*[Na+]^3/K0 = 2*K0^2*K2/(K1*KH) + K0*pCO2*[Na+] + Kw*K0*[Na+]/(K1*KH)

Collecting like powers of [Na+] gives:

{PCO2 + K1*KH*(pCO2)^2/K0}*[Na+]^3 - {K0*pCO2 + Kw*K0/(K1*KH)}*[Na+] - 2*K0^2*K2/(K1*KH) = 0

Plugging in the values for all the equilibrium constants, and assuming pCO2 = 3.8*10^-4 (the present-day value) gives:

3.8*10^-4 * [Na+]^3 - 1.507*10^-4 * [Na+] - 1.23*10^-3 = 0

This is a cubic equation in [Na+]. There are various techniques for solving it. The simplest being to plot the value of the left hand side of the above equation as a function of [Na+] and look for the zeros of the function. I was lazy and used a numerical solver to find the solution.

It turns out that there is only one real solution to this equation, at [Na+]= 1.568, so the equilibrium concentration of Na+ for a saturated solution of NaHCO3 in contact with the modern atmosphere is ~1.568 moles/liter. Because each mole of dissolved NaHCO3 produces one mole of Na+, the molar concentration of NaHCO3 is the same as [Na+]. The number of moles of NaHCO3 in 1 mL of this solution is 1/1000'th of the amount in 1 liter, so there are 1.568*10^-3 moles of NaHCO3 in 1mL of saturated solution.

If instead of the modern value of pCO2= 3.8*10^-4, we use the pre-industrialization value of 3*10^-4, then we have that:

3.0*10^-4 * [Na+]^3 - 1.19*10^-4 * [Na+] - 1.23*10^-3 = 0

This equation again has only one real solution, at [Na+] = 1.683 mol/L. In 1900, one would have measured a saturation concentration of NaHCO3 that was 100*(1.683 - 1.568)/1.568 = 7.33% higher than one would today!

Answer 2

Saturated Sodium Bicarbonate Solution

Answer 3

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Answer 4

It is 4.4643 x 10 -5 mol. of NaHCO3