find the ph of 0.100M NaC2H3O2 solution.
do i just write an equation
and solve for x
NaC2H3O2 +H2O= Na2O + HC2H3O2-
0.100 0 0
so x^2/0.100-x =1.8 X10^-5
x=1.3X10^-3
then do ph=-log(1.3X10-3)
ph=2.9
is this right?
To find the pH of a solution of NaC2H3O2, you need to consider the dissociation of the compound.
NaC2H3O2 dissociates in water as follows:
NaC2H3O2 (aq) → Na+ (aq) + C2H3O2- (aq)
The C2H3O2- ion can act as a weak base, reacting with water to produce OH- ions:
C2H3O2- (aq) + H2O (l) ↔ HC2H3O2 (aq) + OH- (aq)
To determine the pH, you need to find the concentration of the OH- ions. We can consider the initial concentration of NaC2H3O2 (0.100 M) to be equal to the initial concentration of C2H3O2- ion.
Let's denote the concentration of OH- ions as x.
The initial concentration of HC2H3O2 can be considered negligible, so the equilibrium expression for the reaction between C2H3O2- and water is:
x^2 / (0.100 - x) = Kw / [C2H3O2-] = 1.0 × 10^-14 / 0.100
Simplifying the equation gives:
x^2 / (0.100 - x) = 1.0 × 10^-13
Solving this equation, you find that x ≈ 1.0 × 10^-7 M.
As the solution is the result of the reaction of a weak base, the pH can be determined using the formula: pH = -log[H+]
Since [H+] = 1.0 × 10^-7 M, the pH is approximately 7.
Therefore, the correct pH of the 0.100 M NaC2H3O2 solution is 7, indicating the solution is neutral.
To find the pH of a solution of NaC2H3O2, you need to consider the hydrolysis reaction of the acetate ion (C2H3O2-) in water. Here is the correct equation for the reaction:
CH3COO- + H2O ⇌ CH3COOH + OH-
The initial concentration of NaC2H3O2 is 0.100 M, which means the concentration of CH3COO- is also 0.100 M.
Using the equation for the equilibrium constant, you can set up the following expression:
Kw = [CH3COOH][OH-]/[CH3COO-]
Since Kw is a constant value (1.8 x 10^-14 at 25°C), you can rearrange the equation:
[OH-] = (Kw/[CH3COO-])/[CH3COOH]
Substituting the known values:
[OH-] = (1.8 x 10^-14)/(0.100)
[OH-] = 1.8 x 10^-13
Now, you need to find the pOH of the solution using the concentration of OH-:
pOH = -log10([OH-])
pOH = -log10(1.8 x 10^-13)
pOH ≈ 12.74
Finally, you can find the pH using the equation:
pH = 14 - pOH
pH = 14 - 12.74
pH ≈ 1.26
So, the correct pH of the 0.100 M NaC2H3O2 solution is approximately 1.26.