If 20.0 mL of 0.42 M acetic acid (CH
3COOH) is added to 10.0
mL of 0.42 M sodium acetate (CH
3COONa) solution, what is the
pH of the resulting solution? (K
= 1.8 x 10-5)
I notice that your Ka is missing units is this deliberate???
start from the equation
HAc -> H+ + Ac-
so
Ka=[H+][Ac-]/[HAc]
at the start concentration acid
20 ml x 0.42 M/30 ml
=0.28 M
at the start concentration acetate
10 ml x 0.42 M/30 ml
=0.14 M
if H+ concentration at equilibrium = x
at equilibrium
acid concentration
0.28 M - x
acetate concentration
0.14 M +x
substitute
Ka=[H+][Ac-]/[HAc]
Ka=[x][0.14 + x]/[0.28-x]=1.8 x 10-5
solve the resulting quadratic for x
or
assume x is small wrt 0.14
[x][0.14]/[0.28]=1.8 x 10-5
find x.
and pH=-log(x)
Great, thanks so much! :)
And yes Ka doesn't have any units since K is an equilibrium constant.
To find the pH of the resulting solution, we need to calculate the concentration of hydronium ions ([H3O+]). We can do this by using the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acidic component and the ratio of the concentrations of the conjugate acid-base pair.
The acetic acid and sodium acetate form a conjugate acid-base pair with acetic acid being the acid and acetate being its conjugate base. The pKa of acetic acid is given as 4.74 in the reference source for the Henderson-Hasselbalch equation.
First, let's calculate the concentration of acetate ([CH3COO-]) in the solution:
Using the formula: Moles = Molarity x Volume (in L),
Moles of CH3COO- = 0.42 mol/L x 0.01 L = 0.0042 moles
Next, let's calculate the concentration of acetic acid ([CH3COOH]) in the solution:
Moles of CH3COOH = 0.42 mol/L x 0.02 L = 0.0084 moles
Now, let's calculate the ratio of [CH3COOH]/[CH3COO-]:
Ratio = [CH3COOH]/[CH3COO-] = 0.0084 moles / 0.0042 moles = 2.0
Now, we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log10([CH3COOH]/[CH3COO-])
pH = 4.74 + log10(2.0)
Using logarithmic properties, we can calculate the logarithm:
pH = 4.74 + 0.301
Finally, after adding the numbers:
pH = 5.041
Therefore, the pH of the resulting solution is approximately 5.041.