Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 2.1 x 10-2 Ka (HX-) = 7.3 x 10-8
Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 9 x 10-5 Ka (HX-) = 5.3 x 10-11
#1 I believe is just the pH of the salt acting as an acid.
..............HX^- ==> H^+ + X^2-
initial.......0.1......0......0
change........-x.......x.......x
equil.......0.1-x......x.......x
Set up k1 and solve for x. You will not be able to solve without using a quadratic equation.
4). This one looks amazingly like NaHCO3 for which (H^+) = sqrt(k1k2)
DrBob222 don't we have to consider the hydrolysis of HX- here?
H2O(l)+ HX-(aq)<==>H2X(aq)+OH-(aq)-->(1)
equ. (0.1-x) x x
(moldm-3)
Kh= [H2X][OH-]/[HX-]
Kh=kw/ka(HX-)
But we are not given the kW value and we can take it as 14 only if the temperature is 25°C.
Let's assume the temperature is 25°C and kW=14,which gives kh= 10^(-14)/[5.3*(10)^-9]= (1/5.3)*10^(-5) moldm-3=[x^2/(0.1-x)]
if we take (0.1-x) is approximately 0.1,we get,
x^2 = 0.19*(10)^-5==> 2*(10)^-5 M
x==> 4.46*(10)^-2
But we've been given the ka of H2X.
H2X<===> H+(aq) + HX-(aq)->(2)
equ. (x-y) y y
(moldm-3)
which gives,
y^2 =9*(10)^-5 moldm-3 * x
y^2 =9*4.46*(10)^-7 M
Don't we have to consider both (1) and (2) like this to find pH
Some corrections should be added!
x^2==>2*(10)^-6 M
x==>1.41*(10)^-3 M
y^2= 9*1.41*(10)^-8 M
y^2=12.69*(10)^-8 M
y ==>3.56*(10)^-4 M
[H+]=> 3.56*(10)^-4 M
To calculate the pH of a salt solution, we need to consider the dissociation of the salt and the presence of acid-base reactions.
First, let's break down the NaHX salt into its constituent components, Na+ and HX-.
HX- is an anion derived from a weak acid, so it will undergo some degree of hydrolysis in water. The hydrolysis reaction can be represented as follows:
HX- + H2O ↔ H2X + OH-
Let's define the initial concentration of NaHX as 0.10 M. Since Na+ is the conjugate base of a strong acid, it won't undergo any chemical reactions and can be considered spectator ion.
Next, let's consider the dissociation of HX-. The acid dissociation constant (Ka) for HX- can be used to determine the concentration of H+ or OH- ions in the solution.
The equilibrium expression for the hydrolysis reaction of HX- can be written as:
HX- + H2O ↔ H2X + OH-
The equilibrium constant for this reaction, Kb (base dissociation constant), is the reciprocal of Ka for HX-. In this case, Kb = 1/Ka.
Now, let's calculate the concentration of H+ ions in the solution to determine the pH.
1. Calculate the concentration of OH- ions using the Kb value:
Kb = [H2X][OH-] / [HX-]
Given:
Kb = 1/Ka(HX-) = 1 / (5.3 x 10^-11) = 1.89 x 10^10
Assuming the hydroxide ion (OH-) concentration to be x, the concentration of H2X (HX-) and the concentration of OH- are both x.
Therefore, [H2X] = [OH-] = x.
Substituting these values into the Kb expression:
Kb = (x)(x) / (0.10 - x)
Rearranging the expression to solve for x:
x^2 / (0.10 - x) = 1.89 x 10^10
2. Solve the quadratic equation:
x^2 - (0.10 - x)(1.89 x 10^10) = 0
Solving this equation will give us the value of x, which represents the concentration of OH- ions in the solution. We can then use this information to calculate the concentration of H+ ions.
3. Once the concentration of OH- ions is determined, we can calculate the concentration of H+ ions:
[H+] = Kw / [OH-]
where Kw is the ion product of water (1.0 x 10^-14 at 25°C).
4. Finally, the pH is calculated by taking the negative logarithm (base 10) of the H+ concentration:
pH = -log[H+]
By following these steps and solving the equations, you should be able to determine the pH of the 0.10 M NaHX salt solution.