Midterm study guide

Question 1

Why are mmol/kg used as units for marine chemistry, instead of mmol/L?

Answer 1

The ocean is deep, and water is slightly compressible. So 1.0 L of water at the surface (sea level) would be slightly smaller in volume at depth, where pressure is higher. So we use mass of seawater, rather than volume, as units for the concentration of solutes in seawater

Question 2

How do we know that the salt composition of the ocean is similar today to that found
900,000,000 years ago?

Answer 2

The composition, and layering, of evaporites made from ancient oceans have the same sequence of salts we would expect from our present ocean. In other words, calcium carbonate (CaCO3)
precipitates out first, then CaSO4, then NaCl, then MgSO4, and so on.
If CaCO3 was not found at the bottom of evaporite deposits, then the ocean chemistry would not have been the same 900,000,000 years ago as it is today. For example, if you found the very insoluble AgCl at the bottom, then you would know that there was a whole lot more Ag+ in ancient seawater than there is
today (practically none)

Question 3

The d18O(l) value of surface sea water at the equator is about 0.00‰.
approximate d18O
value of water vapor that evaporates from this sea water is -13.00%.
a) Write the chemical equation that describes this process
b) What is the equilibrium constant (K or alpha) for this isotopic process?

Answer 3

a. H2O(l) -> H2O(v)
b. Using the equation alpha = K = ((dproducts – dreactants)/1000)+1 = ((-13.0-0.0)/1000)+1 = 0.987
* (α-1)*1000 = dproducts – dreactants (a very useful eqn!)
The fact that the K or alpha is less than 1.0 means that the d18O of the product (vapor phase) has to be
lower than the d18O of the reactant (liquid phase).

Question 4

The [K+] and [Cl-] in open ocean seawater at salinity of 35 psu are 10.0 and 550. mmol/kg-sw,
respectively.
a) What are the [K+] and [Cl-] in open ocean seawater that has a salinity of 37 psu?
b) Where in the ocean would you expect to find a salinity of 37 psu? Give a location (latitude and longitude) and depth.

Answer 4

a. (37/35) = 1.0571 so K+ = 10.0 mmol/kg-sw(1.0571) = 10.6 mmol/kg-sw
Cl-+ = 550. mmol/kg-sw(1.0571) = 581 mmol/kg-sw
b. near surface (upper 100m) in the north or south Atlantic (e.g. 32N, 64W off Bermuda)

Question 5

4. a) What is the concentration of Ar gas (in mol/kg) in surface seawater at 20°C? Assume the Henry’s
   Law constant for Ar at this temperature is 0.0014 mol/kg/atm, and the fraction (pAr) of Ar in air is
   0.0093? Assume a total air pressure of 1.0 atm. Assume 1kg seawater =1000cm3
   b) What is the Feq (flux) of Ar into the seawater (in mmol/cm2/sec) at equilibrium (input of gas from air
   to sea = output of gas from sea to air)? The diffusion constant (D) of Ar in seawater is 2.0 x 10-5 cm2/sec,
   and the stagnant boundary layer thickness is 40 μm. There are 100cm/m, and 1000kg/m3.

Answer 5

a. Caq = KhpAr = (0.0014mol/kg/atm)0.0093atm=0.000013mol/kg
= 1.3x10-5mol/kg
b. Feq= (D/z)(Caq)=[(2x10-5 cm2/sec) / (40x10-4cm)] (1.3x10-5mol/kg)
=(2.0x10-5 cm2/s) / (40x10-4cm) * (1.3x10-5mol/1000cm3)*(1000mmol/mol)
=6.5x10-8mmol/cm2/sec

Question 6

7. Consider the following equation for the dissolution (dissolving) of calcite in water.
   CaCO3(s) + CO2(aq) + H2O <===> Ca 2+(aq) + HCO3 –(aq)
   The equilibrium constant for this reaction is Keq = 5.4 x 10-9
   a) Balance the equation above.
   b) What is the ∆G for this reaction at 25°C? Assume R=.0082 kj/mol/K. Show your work and units
   (c) Is this reaction favored/spontaneous (go toward products)? Why?

Answer 6

a. CaCO3(s) + CO2(aq) + H2O <===> Ca 2+(aq) + 2HCO3 –(aq)
b. ∆G = -RTlnK = -(0.0082 kj/mol/K) * (298 K)* ln(5.4 x 10-9) = 46 kj/mol
c. No it is not favored because ∆G is positive. You can also tell it is not favored because the
K is so much lower than 1. Keq = 5.4 x 10-9