Question

River water between Back Falls and Angel Falls contains 0.0769 mg Al3+ per L and has a pH of 8.20. Total carbonates are 0.005 mol L-1 .

(a) [5 marks] Calculate the concentrations of HCO3 - (aq) and CO3 2- (aq) (in mol L-1 ). For H2CO3(aq), the hydrolysis constants are Ka1 = 4.45 × 10-7 and Ka1 = 4.69 × 10-11

(b) [5 marks] Calculate the alkalinity

Answer 1

The alkalinity is a water characteristics that quantifies the capacity of water to accept hydrogen ion (H⁺). Alkalinity of natural water is mainly due to the presence of two forms of the carbonate ions that acts as a buffer system. Alkalinity is a sum of ion concentrations

Alkalinity = (Hco₃-)+ 2*(co₃(-2))+(OH^-)-(H⁺)

since HCO₃^- and CO₃(-2) carbonate ions exits in equilibrium , their fractions depends on PH of the solutions . PH is the way to express the concentration of H⁺ such sa PH = -log 10 (H⁺)

K1 and K2 are the known constants that describe equilibria of carbonate ions

K1 = 4045*10^{-7 ,} K2 = 4.69*10^-11

(HCO₃(-)) = (4.45*10^-7* 1*10^-9)/((1*10^-9*1*10^-9)+4.45*10^-7*10*10^-9+(4.45*10^-7*4.69*10^-11))

= 4.45*10^-16/(1*10^-18+4.45*10^-16+2.087*10^-7)

=4.45*10^-16/4.4669*10^-16

= 0.953

(CO₃(-2) = (4.45*10^-7*4.469*10^-11)/(1*10^-9*1*10^-9*)+(4.45*10^-7*1*10^-9)+(4.45*10^-7*4.469*10^-11)

= 2.087*10^-17/4.669*10^-16

^{= 0.045}

^{(HCO₃)= 0.0953*0.10.1m}

   ^=0.0953m

^{(CO3(-2)=0.045*0.1m}

   ^{= 0.0045m}

^{(ALkalinity =( HCO3-)+2(CO3(-2))+(OH-)}

^{ALkalinity = 0.0953+2*0.0045*0.00001}

^{= 0.1043(eq/L)}