What is the pH of a 0.1 M NaOCl solution (Ka of HOCl is 2.95 x10-8)
I did:
x squared/.10= 2.95x10-8
x squared= 2.95x10-9
x= 5.43x10-5
-log(5.43x10-5)=
4.27 pH
However, the options I'm given are either
a.) 3.74 b.) 6.47 c.) 8.23 d.) 10.27
What steps am I doing wrong?
You are treating this as an acid perhaps. NaOCl is a salt and the pH is determined by the hydrolysis of the salt; i.e., Kb of OCl^-.
........OCl^- + HOH ==> HOCl + OH^-
I.......0.1M..............0.....0
C.......-x................x.....x
E......0.1-x..............x.....x
Kb for OCl^- = (Kw/Ka for HOCL) = (x)(x)/(0.1-x) = ?
Solve for x = (OH^-), convert to pOH then to pH. My calculator is on the blink and I can't calculate that; however, it appears that c and d are in the right ballpark.
Oh, pH calculations can sometimes be confusing, but don't worry, I'm here to help! It seems like you made a small error in your calculation. Let's go through the steps again together.
The given solution is NaOCl, which can act as a weak base. To find the pH, we need to consider the hydrolysis reaction of NaOCl:
NaOCl + H2O ⇌ NaOH + HOCl
Since Ka of HOCl is given (2.95 x 10^-8), we know that HOCl will partially ionize in water. The concentration of OH- formed will be equal to the concentration of NaOCl that hydrolyzes.
Let's represent the concentration of OH- formed by 'x'.
From the balanced equation, we can see that the initial concentration of NaOCl is 0.1 M, and the initial concentration of OH- is 0 M. After the reaction, 'x' moles of NaOCl will hydrolyze, resulting in 'x' moles of OH- being produced.
To set up the equilibrium expression for the reaction, we write:
x^2 / (0.1 - x) = 2.95 x 10^-8
Now, here comes a little simplification trick. Since 'x' is much smaller than 0.1, we can approximate the denominator by just using 0.1.
x^2 / 0.1 = 2.95 x 10^-8
Now, let's solve for 'x':
x^2 = 2.95 x 10^-9
x ≈ 5.43 x 10^-5
And now, to find the pH, we take the negative logarithm of 'x':
-pH = log(5.43 x 10^-5)
pH ≈ -log(5.43) - log(10^-5)
Using logarithm properties, we can simplify it further:
pH ≈ -log(5.43) + 5
Calculating this, we get a value close to 2.26.
Hmm, it seems that none of the given options match perfectly. Maybe there was a mistake in the answer choices, or perhaps we made another minor calculation error. It's good to double-check and be cautious when dealing with pH calculations.
Your calculations seem correct, but there might be a mistake in the way you use the logarithm function. The pH of a solution is calculated by taking the negative logarithm (base 10) of the concentration of H+ ions.
In this case, you found that the concentration of H+ ions is 5.43 x 10-5 M. To find the pH, you should take the negative logarithm of this concentration:
-pH = -log(5.43 x 10-5)
Using a calculator, the correct calculation would be:
-pH = -log(5.43 x 10-5) ≈ -(-4.2647) ≈ 4.26
So the pH of the 0.1 M NaOCl solution should be approximately 4.26. None of the options (a, b, c, or d) provided match this value exactly. Perhaps there was a mistake or typo in the options given.
To determine the pH of a solution, you need to take into account the dissociation of the acid or base present. In this case, NaOCl is a salt formed from the reaction between sodium hydroxide (NaOH) and hypochlorous acid (HOCl).
The dissociation equation for the reaction is: HOCl ⇌ H+ + OCl-.
The Ka expression for the dissociation of HOCl can be written as:
Ka = [H+][OCl-] / [HOCl].
Given that the concentration of NaOCl is 0.1 M, the concentration of HOCl is also 0.1 M because the stoichiometry of the reaction is 1:1.
Now, let's solve for the concentration of H+ ions using the given Ka value and the initial concentration of HOCl:
2.95 x 10^-8 = [H+][OCl-] / [HOCl]
2.95 x 10^-8 = [H+][0.1] / [0.1]
2.95 x 10^-8 = [H+]
So, [H+] = 2.95 x 10^-8 M.
To calculate the pH, we need to take the negative logarithm of the concentration of H+ ions:
-pH = -log[H+]
= -log(2.95 x 10^-8)
≈ -(-7.532)
≈ 7.532
Therefore, the pH of the 0.1 M NaOCl solution is approximately 7.532.
From the options you provided, the closest pH value is option b) 6.47.