what is the pH of an aqueous solution that is 0.123M NH4Cl
where did you get m?
[(NH3)(H3O^+)/(NH4^+)] = Kw/kb
(y*y/0.123-y)= 1 x 10^-14/1.8 x 10^-5
I don't know your book value for Kb for NH3 but I'm using 1.8 x 10^-5.
Then operate on the right side only to get
1 x 10^-14/1.8 x 10^-5 = 5.56 x 10^-10, then
[y^2/(0.123-y)] = 5.56 x 10^-10
Since y is small compared to 0.123, we can simplify by making 0.123-y = 0.123, then (or you can solve the quadratic if you wish)
(y^2/0.123)=5.56 x 10^-10
y^2 =5.56 x 10^-10*0.123 = 6.83 x 10^-11 and y = sqrt (...) = 8.27 x 10^-6
Convert that to pH which is approximately 5.10.
my y = 2.25*10^-5
n what do i do
Well, well, well, looks like we have a chemistry question here! Don't worry, I won't make any explosive jokes. The pH of an aqueous solution can be determined by the concentration of hydrogen ions (H+). In this case, NH4Cl is a salt that dissociates into NH4+ and Cl- ions in water. NH4Cl is a weak acid, and its conjugate base, NH4+, can act as a weak acid as well. So, the solution will be slightly acidic. However, without knowing the Ka value, it's tough to give a precise pH value. But, since it's a weak acid, we can say it'll be slightly below 7 on the pH scale. I hope that answers your question without causing too much turbulence!
To find the pH of an aqueous solution, we need to consider the dissociation of the solute and its effect on the concentration of hydrogen ions (H+) in the solution.
In the case of NH4Cl, it is an acidic salt that dissociates in water to form NH4+ and Cl- ions. The NH4+ ion can act as a weak acid, while the Cl- ion is a spectator ion and does not participate in the pH calculation.
The dissociation reaction of NH4+ in water is:
NH4+ + H2O ⇌ NH3 + H3O+
Since NH4+ can donate a proton (H+) to water, it acts as a weak acid, leading to the formation of ammonium (NH3) and hydronium (H3O+) ions.
Now, let's calculate the initial concentration of the NH4+ ion in the solution. Since the initial concentration of the NH4Cl compound is given as 0.123M, the concentration of NH4+ is also 0.123M because NH4+ comes from the dissociation of NH4Cl.
Next, we can assume that the dissociation of NH4+ is small, so we consider it negligible compared to the initial concentration of NH4+ (0.123M). Therefore, we can approximate the concentration of NH4+ as the initial concentration of NH4+ minus the concentration of H3O+ formed.
Since NH4+ and H3O+ have a 1:1 stoichiometric ratio in the reaction, the concentration of H3O+ is equal to the concentration of NH4+ that dissociates. The concentration of NH4+ that dissociates can be calculated using the acidity constant of NH4+, abbreviated as Ka(NH4+), which is equal to 5.6 x 10^-10 at 25°C.
Now, with these values, we can set up an equation to calculate the concentration of H3O+:
Ka(NH4+) = [NH3][H3O+] / [NH4+]
5.6 x 10^-10 = (x)(x) / (0.123 - x)
Simplifying the equation by assuming x is much smaller than 0.123, we have:
5.6 x 10^-10 ≈ x^2 / 0.123
Rearranging the equation, we get:
x ≈ √(5.6 x 10^-10 * 0.123)
Calculating this expression, we find that x ≈ 2.35 x 10^-6
Since x represents the concentration of H3O+ ions, which is equivalent to the concentration of H+ ions in water, the pH of the solution is equal to the negative logarithm (base 10) of the concentration of H+:
pH = -log[H+]
pH ≈ -log(2.35 x 10^-6)
Using a calculator, we find that the pH is approximately 5.63.
Therefore, the pH of the aqueous solution that is 0.123M NH4Cl is approximately 5.63.
NH4Cl is a salt. It hydrolyzes, (reacts with water)(actually, the NH4^+ hydrolyzes) to give
NH4^+ + HOH ==> NH3 + H3O^+
Then Ka, since the NH4^+ is acting as an acid = (NH3)(H3O^+)/(NH4^+)
(NH3) = y
(H3O^+) = y
(NH4^+) = 0.123-y
solve for y.
You don't know Ka but you can calculate it since KaKb=Kw.