Question

a) (7.5 points) Calculate the pH of a sample of river water in which [HCO, is 1.4 x 104M and the concentration of dissolved CO2 (which is maintained constant by equilibration with atmospheric CO2) is 14x 103M. (pKa for the Co,/Hco, Ico,2" system is 6.37.)

Please show all work

Answer 1

a)

Assume

CO2 + H2O = H2CO3

H2CO3 = HCO3- + H+

then

pH = pKa + log(HCO3-/H2CO3)

pH = 6.37 + log((1.4*10^-4) / (1.4*10^-5))

pH = 7.37

b)

mmol of HNO3 = MV = (10^-3)(10) = 0.01

H+ + HCO3- = H2CO3

mmol of HCO3- initially = (1.4*10^-4)(10^3)= 0.14

mmol of H2CO3 initially = (1.4*10^-5)(10^3) =0.014

after 0.01 mmol addition

HCO3- left = 0.14-0.01 = 0.13

H2CO3 = 0.014+0.01 = 0.024

pH = 6.37 + log(0.13/0.024) = 7.103

c)

initial pH ofw ater = 7

pH change if:

mmol of H+ = MV = (10^-3)(10) = 0.01

Vtotal = 1000+10 = 1010 mL

[H+] = 0.01/1010 = 0.0000099

pH = -log(0.0000099) = 5.00

pH change = -2