Question

180Q2

Calculate the concentrations of all species present (H2CO3, HCO−3, CO2−3, H3O+ and OH−) in 0.019 M H2CO3. pH is 4.04

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Answer 1

Step 1: In this step, write the chemical equation: Means; [HCO; ][ Ho]=x=9.14x10° H,CO, (aq)+H20(1) HCO3 (aq)+Hot (aq) HCO; (aq) +H20 (1) = 00- (aq) +H,0+ (aq) and expression for equilibrium constant (K, and ka): ( Heo J[ Hạo] [H.CO; ] [co [H0] ke (HCO; ] Step III: Let us if the value of co- concentration is y than Ka be expressed as follows: [HCO; [H:0"]=x=9.14*10-s [co H0] ** [HCO; ] 4.7x10-11, yx9.14x10-5 and K = 4.4x10- Ko = 4.7x10-11 9.14x10- v-4.7x10-4x214x10- 914210 y=4.7x10-11 Therefore the concentrations of H+, HCO3-, and CO32- in a 0.019M H2CO3 solution are following: [HCO ][ Ho]=x= 9.14x10° [co- ] = y = 4.7x10-11 Step II: The stoichiometry of the chemical equation explains that when hco; forms, at the same time same concentration of H,0* ion also forms. Let us if the value of both concentrations is x than K, be expressed as follows: - [HCO ][ Ho] [H+] [H3O+]=1.0*10^-14 [OH-]=1.0*10^-14 / 9.14*10^-5 = 1.09*10^-10 K [H.CO 4.4x10-7 = XXX 0.019 x = 4.4x10-7x0.025 = 8.36x109 x=9.14x10-5