Calculate the pH of a 0.5 M solution of NaOCl. The pKa of HOCl is 7.5. (A: 10.1)
I know the answer but I don't know how to solve this problem.
I made an ICE chart
HOCl + H2O <==> OCl + OH-
I 0.5......-.................0........0
E 0.5 - x.................x..........x
Ka = x^2/0.5-x
Ka = 10^-pka = 10^-7.5 = 3.16e-8. then use that value to sub in the Ka formula to find x.
The ICE chart is OK but you don't have HOCl. You have NaOCl.
.............OCl^- + HOH --> HOCl + OH^-
I..............0.5......................0..............0
E..............0.5-x..................x..............x
Kb for OCl^- = Kw/Ka for HOCl = (HOCl)(OH^-)/(OCl^-)
Plug in the E line solve for x = (OH^-), then convert to pOH then pH.
Post your work if you get stuck.
The equation you wrote with HOCl is for the ionization of HOCl. That would be HOCl + H2O ==> H3O^+ + OCl^- but that isn't what you want for the hydrolysis of anion part of the NaOCl salt.
To calculate the pH of a solution of NaOCl, we need to consider the dissociation of NaOCl in water. NaOCl will dissociate into Na+ and OCl- ions.
The OCl- ion can react with water to form HOCl and OH- ions, which is an acidic reaction. The equilibrium constant for this reaction is given by:
Ka = [HOCl][OH-] / [OCl-]
The pKa of HOCl is given as 7.5, which means that the equilibrium constant Ka can be calculated as follows:
Ka = 10^(-pKa)
Substituting the pKa value, we find:
Ka = 10^(-7.5)
Next, we need to consider the dissociation of water itself. Water can dissociate into H+ and OH- ions:
H2O ⇌ H+ + OH-
At equilibrium, the concentration of H+ ions can be approximated by [H+], which is equal to the concentration of OH- ions in pure water or 1.0 x 10^(-14) M.
Now, we can write the equilibrium expression for NaOCl dissociating in water as follows:
NaOCl ⇌ Na+ + OCl-
Since NaOCl is a strong electrolyte, it will dissociate completely, which means that the concentration of Na+ ions is equal to the initial concentration of NaOCl (0.5 M), and the concentration of OCl- ions is also 0.5 M.
Considering the reaction between OCl- and water, we can set up the equilibrium expression as follows:
Ka = [HOCl][OH-] / [OCl-]
We know that [OH-] = 1.0 x 10^(-14) M, and let's assume the concentration of HOCl as x M.
Substituting these values into the equilibrium expression, we have:
10^(-7.5) = x * (1.0 x 10^(-14)) / (0.5)
Simplifying this equation, we get:
10^(-7.5) = x * (2.0 x 10^(-15))
Dividing both sides of the equation by (2.0 x 10^(-15)), we have:
x = (10^(-7.5)) / (2.0 x 10^(-15))
Simplifying this expression, we get:
x = 5.0 x 10^(-8.5)
Finally, we can calculate the pH of the solution by taking the negative logarithm of the concentration of H+ ions, which in this case is equal to the concentration of HOCl:
pH = -log[H+]
= -log(x)
= -log(5.0 x 10^(-8.5))
= 10.1
Therefore, the pH of a 0.5 M solution of NaOCl is 10.1.
To solve this problem, you can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base. The equation is as follows:
pH = pKa + log([A-]/[HA])
In this case, the acid (HA) is HOCl and the conjugate base (A-) is OCl-.
Given that the pKa of HOCl is 7.5, we can substitute the values in the equation:
pH = 7.5 + log([OCl-]/[HOCl])
Next, you need to determine the ratio of [OCl-] to [HOCl]. Since NaOCl is a strong electrolyte, it fully dissociates in water, giving one Na+ ion and one OCl- ion. Therefore, the concentration of OCl- is 0.5 M.
The concentration of HOCl can be calculated using the principle of electroneutrality. Since NaOCl is a salt, the concentration of HOCl can be assumed to be the same as the concentration of OCl-. Therefore, [HOCl] = 0.5 M.
Substituting these values into the Henderson-Hasselbalch equation:
pH = 7.5 + log(0.5/0.5)
The log of 1 is 0, so:
pH = 7.5 + 0
pH = 7.5
Therefore, the pH of a 0.5 M solution of NaOCl is 7.5.