Question

A water is in equilibrum with the atmosphere (P_co² = 10^-3.5 atm) at 25^oC, and has a pH of 8.1. No other weak acid or weak bases are present. Ignore activity corrections. Assume K_H = 10^-3.5 moles/(L-atm). a) What is the alkalinity of this water in meq/L. b) What volume (in mL) of 0.4 M H₂SO₄ is needed to decrease the pH of 1 liter of this water to the alkalinity titration endpoint? ignore dilution effect

Answer 1

Alkalinity of the water = (K_H * P_CO2 )/[H⁺] where K_H = 10^-3.5 moles/L-atm and P_CO2 =10^-3.5 atm and pH = 8.1

pH -log[H⁺] ,therefore [H⁺] = antilog (-8.1) = 7.94 *10^-9

Alkalinity of the water = (K_H * P_CO2 )/[H⁺] = (10^-3.5 moles/L-atm* 10^-3.5 atm)/7.94*10^-9

Alkalinity of the water = (9.99*10^-8 moles/L)/7.94*10^-9

Alkalinity of the water = 12.58 moles/L into meq/L = 12.58moles/L*((18 eq/moles)/1)*(1/1000)

a) Alkalinity of the water = 0.226 meq/L = 10^-0.64 meq/L

b) 0.4 M sulphuric acid

M_H2SO4V_H2SO4 = M_H2OV_H2O

0.4 * V_H2SO4 = 7.94*10^-9 * 1000mL (1L =1000 mL)

V_H2SO4 = 1.98 *10^-5 mL is required to decrease the pH of 1L of this water to the alkalinity titration end point