Calculate the H3O+ concentration and pH of a carbonated beverage that is .10 M in dissolved CO2- (Essentially all of the H3O+ comes from the first stage of dissociation, CO2 + 2H2O = H3O^+ + HCO3-, for which K1 = 4.4 x 10^-7)
H2CO3 ==> H^+ + HCO3^-
k1 = 4.4E-7 = (x)(x)/(0.1-x)
Solve for x, convert to pH.
To determine the H3O+ concentration and pH of a carbonated beverage, we will use the given information that the dissolved CO2- concentration is 0.10 M and the first dissociation constant (K1) is 4.4 x 10^-7.
Let's start by writing the reaction equation for the dissociation of CO2 in water:
CO2 + 2H2O → H3O+ + HCO3-
The dissociation constant (K1) is defined as the ratio of the concentrations of the products (H3O+ and HCO3-) to the concentration of the reactant (CO2). Therefore, we can write the equilibrium expression as follows:
K1 = [H3O+][HCO3-] / [CO2]
Since the concentration of CO2 is given as 0.10 M, we can substitute this value into the equation:
4.4 x 10^-7 = [H3O+][HCO3-] / 0.10
Now, let's assume that x is the concentration of H3O+ and HCO3-. As we can see from the equation, 1 mole of CO2 forms 1 mole of H3O+ and 1 mole of HCO3-. Therefore, the concentration of H3O+ will be equal to the concentration of HCO3-.
So, [H3O+] = [HCO3-] = x
Substituting this into the equation:
4.4 x 10^-7 = (x)(x) / 0.10
Rearranging the equation gives:
4.4 x 10^-7 = x^2 / 0.10
Solving for x using algebraic methods or a calculator, we find:
x ≈ 6.63 x 10^-4 M
So, the H3O+ concentration in the carbonated beverage is approximately 6.63 x 10^-4 M.
To calculate the pH, we can use the equation:
pH = -log[H3O+]
For our case:
pH = -log(6.63 x 10^-4)
Using a calculator, we find:
pH ≈ 3.18
Therefore, the H3O+ concentration of the carbonated beverage is approximately 6.63 x 10^-4 M, and its pH is about 3.18.