Calculate the molar solubilty of CaC2O4 in a solution that has been buffered to a pH of 4.00 Remember to account for reactions involving H2C2O4 acting as an acid in water. To account for the buffer, simply assume that H+ = 1.00X 10^-4
CaC2O4 is more soluble in the acidic buffered solution that it would normally be because of the formation of the bioxalate (hydrogen oxalate) ion.
CaC2O4 ==> Ca^+2 + C2O4^=
And H^+ + C2O4^= ==> HC2O4^-
And H^+ + HC2O4^- ==> H2C2O4
Look up Ksp for CaC2O4.
(1) Ksp = (Ca^+2)(C2O4^=) = ??
(2) Solubility = (Ca^+2) = (C2O4^=) + (HC2O4^-) + (H2C2O4)
(3) k1 H2C2O4 = you can do this.
(4) k2 H2C2O4 = you can do this.
First, the (H2C2O4) is quite small because the hydrolysis of HC2O4^- to form H2C2O4 is of the order of 10^-12 so we can neglect that.
Use k2 (equation 4) and plug in 1 x 10^-4 for H^+ and solve for (C2O4^=)/(HC2O4^-) = xx
and rearrange that to solve for (HC2O4^-) in terms of C2O4^=,
Now use equation 2 to obtain (Ca^+2) in terms of oxalate and rearrange that to solve for oxalate in terms of Ca^+2.
Finally, use equation 1 (Ksp) to plug in for Ca^+2 and for C2O4^= (in terms of Ca^+2). Solve for the only unknown of Ca^+2.
I have an answer of 8.09 x 10^-5 M but check my thinking and check my arithmetic. Post your work if you get stuck.
I would round the 8.09 to 8.1 x 10^-5 since the Ksp I found was to only two significant figures.
To calculate the molar solubility of CaC2O4 in a solution buffered to a pH of 4.00, we need to consider the balanced equation for the dissociation of CaC2O4 and the dissociation of H2C2O4.
The balanced equation for the dissociation of CaC2O4 is:
CaC2O4(s) ⇌ Ca2+(aq) + C2O42-(aq)
The balanced equation for the dissociation of H2C2O4 is:
H2C2O4 ⇌ H+(aq) + HC2O4-(aq)
Since the solution is buffered, the concentration of H+ is given as 1.00 x 10^-4 M. This means that [H+] = 1.00 x 10^-4 M and [HC2O4-] = 1.00 x 10^-4 M.
Let's assume the molar solubility of CaC2O4 is x.
Using the balanced equation for the dissociation of CaC2O4, we can write the expression for the solubility product constant (Ksp) as:
Ksp = [Ca2+][C2O42-] = x(2x) = 4x^3
Since CaC2O4 dissolves into Ca2+ and C2O42-, the concentration of Ca2+ and C2O42- will be equal to the molar solubility, x.
Using the balanced equation for the dissociation of H2C2O4, we can write the expression for the acid dissociation constant (Ka) as:
Ka = [H+][HC2O4-] / [H2C2O4] = (1.00 x 10^-4)^2 / 1.00 x 10^-4 = 1.00 x 10^-4
Since HC2O4- acts as an acid, the concentration of HC2O4- will be equal to the concentration of H+.
Now, we can write an expression for the concentration of C2O42- in terms of x and [H+]:
[C2O42-] = x + [H+] = x + 1.00 x 10^-4
We can substitute this expression into the solubility product constant expression:
Ksp = (x)(x + 1.00 x 10^-4)^2 = 4x^3
Now, we can solve this cubic equation to find the value of x. However, since solving cubic equations can be complex, we can approximate the x value by assuming that the value of 1.00 x 10^-4 is negligible compared to x. This allows us to simplify the equation:
4x^3 ≈ (x)(x)^2 = x^3
Simplifying further, we get:
4 ≈ x
Therefore, the molar solubility of CaC2O4 in the buffered solution is approximately 4 M.
To calculate the molar solubility of CaC2O4, we need to consider the equilibrium reaction that occurs when CaC2O4 dissolves in water. The balanced equation for this reaction is:
CaC2O4 (s) ⇌ Ca2+ (aq) + C2O4 2- (aq)
To account for the buffer in the solution, we need to consider the dissociation of the acid H2C2O4 (oxalic acid) into H+ ions and C2O4 2- ions. From the given information, we know that the concentration of H+ ions is 1.00 X 10^-4 M (moles per liter).
H2C2O4 (aq) ⇌ 2H+ (aq) + C2O4 2- (aq)
To calculate the molar solubility, we need to determine the concentration of Ca2+ ions at equilibrium. Since CaC2O4 is a sparingly soluble salt, we can assume that its solubility is "x" moles per liter.
Using the stoichiometry of the reaction, we can see that for every one mole of CaC2O4 that dissolves, one mole of Ca2+ ions is produced. Therefore, at equilibrium, the concentration of Ca2+ ions is also "x" M.
Next, we need to consider the equilibrium expression for the solubility product constant (Ksp) for CaC2O4:
Ksp = [Ca2+][C2O4 2-]
Using the equilibrium concentrations we have determined, the expression becomes:
Ksp = x * x = x^2
Since the molar solubility of CaC2O4 is equal to the concentration of Ca2+ ions, we can rewrite the equation as:
Ksp = x^2
Knowing the value of Ksp for CaC2O4 (which you may need to look up), we can solve for the molar solubility:
x^2 = Ksp
Take the square root of both sides:
x = √Ksp
Therefore, the molar solubility of CaC2O4 in the buffered solution can be calculated by taking the square root of the value of Ksp for CaC2O4.