COMMON ION EFFECT
1.) What is the pH of a solution that is 0.30M in HCOOH and 0.52M in HCOOK.
Ka of HCOOH=1.7x10^-4.
2.) Calculate the pH of a buffer system containing 1.0M CH3COOH and 1.0M CH3COONa. Ka of CH3COOH=1.8x10^-5.
1)
I notice that your value for Ka in the question is missing units, were these missing in the original question?
Start from the equation for the equilibrium reaction
HCOOH<->H+ + HCOO-
at start
0.30M
at equilibrium if [H+]=x
0.30-x x x
there is also 0.52M HCOO-
so at equilibrium
0.30-x x x+0.52
Ka=[H+][HCOO-]/[HCOOH]
Ka=(x)(x+0.52)/(0.30-x)=1.7x10^-4
(x)(x+0.52)/(0.30-x)=1.7x10^-4
You can either solve the quadratic or we can say that if x is small with respect 0.52 abd 0.3, we can rewrite the expression as
(x)(0.52)/(0.30)=1.7x10^-4
and find x
pH is then -log (x/mol litre^-1)
To find the pH of a solution using the common ion effect, we need to consider the dissociation of the weak acid in the presence of its conjugate base.
1.) For the first question, we have a solution that is 0.30M in HCOOH (formic acid) and 0.52M in HCOOK (potassium formate). We know the value of Ka for HCOOH is 1.7x10^-4.
First, we need to determine the concentration of HCOOH that dissociates. Let's assume x M of HCOOH dissociates. Then, the concentration of HCOO⁻ (conjugate base) formed would also be x M.
Using the dissociation equation for HCOOH: HCOOH ⇌ H⁺ + HCOO⁻, we set up an ice table:
Initial: 0.30 M 0.52 M 0 M
Change: -x M +x M +x M
Final: 0.30 - x M 0.52 + x M x M
Now, we can write the equation for Ka:
Ka = [H⁺][HCOO⁻] / [HCOOH]
Since the concentration of H⁺ in this case is equal to the concentration of HCOO⁻ (x M), and the concentration of HCOOH is 0.30 - x M, we can substitute these values into the equation:
1.7x10^-4 = (x)(x) / (0.30 - x)
Simplifying the equation:
1.7x10^-4 = x^2 / (0.30 - x)
Approximating the concentration of HCOOH as approximately 0.30 M (since x would be very small), we can solve for x using the quadratic equation or compare the value of x with 0.30 to check our approximation.
Using the quadratic equation, we can rearrange the equation to form:
x^2 + (0.30 - x)(1.7x10^-4) = 0
Solving this equation will give us the value of x, which represents the concentration of H⁺ and HCOO⁻. Once you have the value of x, you can calculate the pH using the formula:
pH = -log[H⁺]
2.) Similarly, for the second question, we have a buffer system consisting of 1.0M CH3COOH (acetic acid) and 1.0M CH3COONa (sodium acetate). The value of Ka for CH3COOH is given as 1.8x10^-5.
We assume x M of CH3COOH dissociates, resulting in the same concentration of CH3COO⁻ (conjugate base) formed. Using the ice table method, we can set up an equation and solve for x.
Again, we can use the equation for Ka to solve for x:
Ka = [H⁺][CH3COO⁻] / [CH3COOH]
Substituting the values, we get:
1.8x10^-5 = (x)(x) / (1.0 - x)
Simplifying the equation, we end up with:
1.8x10^-5 = x^2 / (1.0 - x)
You can solve this equation using the quadratic equation or check the approximation for x as explained earlier. Once you find the value of x, you can calculate the pH using the formula:
pH = -log[H⁺]
Remember to adjust your approximation based on the size of x before calculating the pH.