what is the hydronium-ion concentration in a 0.35m solution of NaCO3? For carbonic acid, Ka1=4.2x10^-7 and Ka2=4.8x10^-11. (Kw=1.0x10^-14)
I'm sure you mean 0.35M solution. Do you know that 0.35M and 0.35m are not the same? This is a hydrolysis problem.
........CO3^2- + HOH ==> HCO3^- + OH^-
I.......0.35M..............0.......0
C.......-x.................x.......x
E.......0.35-x.............x.......x
Kb for CO3^- = (Kw/k2 for H2CO3) = (x)(x)/(0.35-x). Substitute and solve for x = (OH^-) an convert to pH.
Why is the last step is convert to pH? We are trying to find the hydroxide-ion concentration?not the pH
Nvm figured it out
To find the hydronium-ion concentration in a solution of NaCO3, we need to consider the dissociation of carbonic acid (H2CO3) which is formed when NaCO3 dissolves in water.
Let's break down the dissociation of carbonic acid into steps using the given Ka1 and Ka2 values:
H2CO3 ⇌ H+ + HCO3- (Step 1 - Ka1)
HCO3- ⇌ H+ + CO3^2- (Step 2 - Ka2)
From the above reactions, we can see that H2CO3 can donate two protons (H+), so it can dissociate into two steps.
To find the hydronium-ion concentration, we need to find the concentration of H+ in the solution. Let's assume the initial concentration of H2CO3 is 'x'.
Using the given Ka1 value, the equilibrium expression for Step 1 is:
Ka1 = ([H+][HCO3-])/[H2CO3]
4.2x10^-7 = ([H+]x)/x ⟹ [H+] = 4.2x10^-7
Now, using the given Ka2 value, the equilibrium expression for Step 2 is:
Ka2 = ([H+][CO3^2-])/[HCO3-]
4.8x10^-11 = ([H+]^2)/[HCO3-]
4.8x10^-11 = (4.2x10^-7)^2/[HCO3-]
Now, we need to find the concentration of HCO3-. Since NaCO3 is a strong electrolyte, it fully dissociates in water into Na+ and CO3^2-. Therefore, the concentration of CO3^2- is equal to the initial concentration of NaCO3, which is given as 0.35M.
[HCO3-] = [CO3^2-] = 0.35M
Substituting the values into the equation:
4.8x10^-11 = (4.2x10^-7)^2/0.35
[H+]^2 = 0.35 * 4.8x10^-11 / (4.2x10^-7)^2
[H+] = √(0.35 * 4.8x10^-11) / 4.2x10^-7
Now, calculate [H+] using the given equation.
Once you solve this equation, you will obtain the concentration of hydronium ions ([H+]) in the 0.35M NaCO3 solution.