Question

A sample of rainwater is observed to have a pH of 7.4. If only atmospheric CO2 at 390 ppm and limestone dust are present in the atmosphere to alter the pH from a neutral value, and if each raindrop has a volume of 0.02 cm3, what mass of calcium (in ng) is present in each raindrop? (Do not make any simplifications.)

Note: the answer is 59 ng. some say 0.02 cm3 rain drop is equal to 0.0134 L, how did that calculation occur?

Answer 1

A sample of rainwater is observed to have a pH of 7.4. If only atmospheric CO2 at 380 ppm and limestone dust are present in the atmosphere to alter the pH from a neutral value, and if each raindrop has a volume of 0.02 cm3, what mass of calcium is present in each raindrop?

There are following equilibrium for this:

CO₂ (aq) + H₂O     H₂CO₃ (aq)

H₂CO₃ (aq) H⁺ (aq) + HCO₃^- (aq) pKa1 = 6.35

Amount of CO₂ Dissolved in be calculated from Henry 's law ;  

[CO₂ (aq) ] = K_CO2  * P_CO2

(K_CO2  = Henry constant for CO₂ = 3.40*10^-2mol /(L-atm) )

Note : Here we are using 'Henry solubility defined via concentration'.

P_CO2 = 380 * 10^-6 atm (Dalton's partial pressure law)

[CO₂ (aq) ] =  3.40*10^-2mol /(L-atm) * 380 * 10^-6 atm = 1.29 *10^-5 mol/ L

We take ,    [CO₂ (aq) ] =  [ H₂CO₃ ]

H₂CO₃ (aq) H⁺ (aq) + HCO₃^- (aq) pKa1 = 6.35 : Ka1 =10^-6.35

  Ka1 = 10^-6.35=  [ H⁺ ][HCO₃^-  ] / [ H₂CO₃ ]  

since, pH = 7.4

so, we have [ H⁺ ] = 10^-7.4

[HCO₃^-  ] = 10^-6.35 *1.29 *10^-5 mol/ L / 10^-7.4 = 1.45*10^-4 M

Since, only atmospheric CO2 and limestone dust are present in the atmosphere to alter the pH from a neutral value :

we have , [Ca2+] = 1/2 *[HCO₃^-  ] = 7.3 *10^-5 M     (electroneutrality)

so, in rain water, [Ca2+] = 7.3 *10^-5 mol/L

So, mass of Ca in 0.02 cm3 Volume raindrop :

1 L = 1000 cm3

M(Ca) = 40 g/mol

so, : 7.3 *10^-5 mol/L*40 g/mol* 1L/ 1000 cm3 * 0.02 cm3

Thus, mass of Ca in 0.02 cm3 Volume raindrop :5.8*10^-8 g   or 58 ng.

and your note point is not correct

thank you ?

all the best.....