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
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