The Ka values of H2CO3 are shown below.
Ka1 4.6 10-7
Ka2 4.4 10-11
What is the pH of a 0.17 M solution of Na2CO3?
To determine the pH of a 0.17 M solution of Na2CO3, we need to consider the reactions that occur when Na2CO3 dissolves in water.
Na2CO3 is a salt that completely dissociates into its ions in water:
Na2CO3 → 2Na+ + CO32-
The carbonate ion (CO32-) can react with water through two sequential acid-base reactions, resulting in the formation of bicarbonate (HCO3-) and hydroxide (OH-) ions:
CO32- + H2O ⇌ HCO3- + OH-
We can calculate the concentrations of HCO3- and OH- using the following equilibrium expressions:
Ka1 = [HCO3-][OH-] / [CO32-]
Ka2 = [H2O][HCO3-] / [CO32-]
Since the concentration of CO32- ions is equal to the concentration of Na2CO3 (0.17 M), we can substitute the values into the equations:
Ka1 = [HCO3-][OH-] / 0.17
Ka2 = [H2O][HCO3-] / 0.17
Since the concentration of water remains relatively constant, we can assume it to be 1 (significantly higher than the HCO3- concentration). Thus, we can simplify the equations to:
Ka1 = [HCO3-][OH-] / 0.17
Ka2 = [HCO3-] / 0.17
Rearranging the second equation gives us:
[HCO3-] = Ka2 * 0.17
Now, we can substitute this value of [HCO3-] back into the first equation:
Ka1 = [HCO3-][OH-] / 0.17
Ka1 = (Ka2 * 0.17)[OH-] / 0.17
Ka1 = Ka2[OH-]
Simplifying this equation, we find that [OH-] = Ka1 / Ka2.
Now, we can calculate the concentration of OH-:
Ka1 = 4.6 x 10^-7
Ka2 = 4.4 x 10^-11
[OH-] = (4.6 x 10^-7) / (4.4 x 10^-11)
[OH-] = 1.05 x 10^4
Since we have the concentration of OH-, we can calculate the concentration of H+ (from water) using the equation Kw = [H+][OH-]:
Kw = 1.0 x 10^-14
[H+][OH-] = 1.0 x 10^-14
[H+][(4.6 x 10^-7) / (4.4 x 10^-11)] = 1.0 x 10^-14
[H+] = (1.0 x 10^-14) / [(4.6 x 10^-7) / (4.4 x 10^-11)]
Calculating this value, we find that [H+] = 9.57 x 10^-4.
Finally, we can calculate the pH of the solution using the formula: pH = -log[H+]:
pH = -log(9.57 x 10^-4)
pH ≈ 3.02
Therefore, the pH of a 0.17 M solution of Na2CO3 is approximately 3.02.
To determine the pH of a solution of Na2CO3, we need to consider the hydrolysis reaction that occurs when the compound dissolves in water.
Na2CO3 dissociates in water to produce Na+ and CO32- ions. The CO32- ion can react with water to form HCO3- and OH- ions:
CO32- + H2O ⇌ HCO3- + OH-
Since this reaction involves the formation of hydroxide ions (OH-), we need to consider the contribution of OH- to the overall pH of the solution.
Na2CO3 is a salt of a weak acid, H2CO3. The equilibrium constant for the first dissociation step (Ka1) can be used to calculate the concentration of HCO3- ions:
H2CO3 ⇌ H+ + HCO3-
From the given Ka values,
Ka1 = [H+][HCO3-] / [H2CO3]
We can rearrange this equation to solve for [H+]:
[H+] = Ka1 * [H2CO3] / [HCO3-]
Since Na2CO3 dissociates to produce twice as many HCO3- ions as H2CO3, the concentration of HCO3- can be calculated as 2 times the initial concentration of Na2CO3:
[HCO3-] = 2 * [Na2CO3]
Now, let's substitute the given values into the equation. We have:
[H+] = (4.6 * 10^-7) * (0.17) / (2 * 0.17)
= 3.91 * 10^-8 M
To find the pOH, we can use the equation:
pOH = -log10 [OH-]
Since the concentration of OH- is equal to the concentration of [H+] due to the stoichiometry of the hydrolysis reaction, we have:
pOH = -log10 (3.91 * 10^-8)
= 7.41
Finally, to calculate the pH, we use the equation:
pH = 14 - pOH
= 14 - 7.41
= 6.59
Therefore, the pH of a 0.17 M solution of Na2CO3 is approximately 6.59.
Write the hydrolysis equation for carbonate ion.
CO3^-2 + HOH ==> HCO3^- + OH^-
Now set up an ICE chart and substitute into the Kb expression.
Kb = (Kw/k2) = (HCO3^-)(OH^-)/(CO3^-2)
(HCO3^-)= (OH^-) = x
(CO3^-2) = 0.17
You know Kw and k2.
Solve for x and convert to pH.
Post your work if you get stuck.