Calculate pH of a 1.0 mol L -1 boric acid solution.
(Ka(H3BO3) = 5.8 x 10 power -10
Please solve this problem. My computer does not have the power symbol.
Can you please show maths and steps.
Is mollar mass required of the above.
Is -log and constant above used.
Thanks
H3BO3 ==> H^+ + H2BO3^-
Set up an ICE chart, substitute into Ka expression, and solve for H^+. Then use pH = -log(H^+) to convert to pH.
At equilibrium, you should have this substituted into Ka.
5.8 x 10^-10=(H^+)(H2BO3^-)/(H3BO3)
5.8 x 10^-10 = (x)(x)/(1-x)
Solve for x, which is H^+, then convert to pH.
Well, well, well, let's get cracking on this pH calculation! And don't worry, I've got your back. No power symbol? No problemo!
First, let's establish the equation for the dissociation of boric acid (H3BO3) in water:
H3BO3 ⇌ H+ + H2BO3-
Now, we need to consider that the Ka value given is for the equation:
H3BO3 ⇌ H+ + H2BO3-
The Ka expression is:
Ka = [H+][H2BO3-]/[H3BO3]
Since we have a 1.0 mol L-1 boric acid solution, we can assume that the initial concentration of H3BO3 is also 1.0 mol L-1.
Let x be the concentration of H+ and H2BO3-. Since the initial concentration of H3BO3 is equal to the concentration of H+ and H2BO3-, the change in concentration for both H+ and H2BO3- is x.
Therefore, the equilibrium concentrations can be expressed as:
[H+] = x
[H2BO3-] = x
And the concentration of H3BO3 can be expressed as:
[H3BO3] = 1.0 - x
Now let's substitute these values into the Ka expression:
Ka = (x)(x)/(1.0 - x)
Given that Ka = 5.8 x 10^-10, we can rearrange the equation:
5.8 x 10^-10 = (x)(x)/(1.0 - x)
Simplifying further, we get:
5.8 x 10^-10 = (x^2)/(1.0 - x)
Now here comes the disappointing bit. We can't solve this equation analytically. We'll have to assume that the value of x is very small compared to 1.0 and ignore it in the denominator (1.0 - x). Trust me, this approximation works just fine.
So, we can rewrite the equation as:
5.8 x 10^-10 ≈ (x^2)/(1.0)
Now, we can simplify further to:
5.8 x 10^-10 ≈ x^2
To solve for x, we'll take the square root of both sides:
x ≈ sqrt(5.8 x 10^-10)
Now, shovel this into a calculator, and you'll find that x ≈ 7.64 x 10^-6.
And voila! We've got the concentration of H+ and H2BO3-. Since the concentration of H+ is equal to x, the concentration of H2BO3- is also x.
To calculate the pH, we use the equation:
pH = -log[H+]
Since the concentration of H+ is x, we can now calculate the pH:
pH ≈ -log(7.64 x 10^-6)
And this, my friend, concludes our math adventure. Just plug this into a calculator, and you'll find the pH of your 1.0 mol L-1 boric acid solution.
Remember, laughter is the best pH balancing agent!
To calculate the pH of a boric acid solution, we first need to understand the dissociation of boric acid. When boric acid (H3BO3) dissolves in water, it donates a proton (H+) to the water molecule, resulting in the formation of the borate ion (B(OH)4-).
The dissociation reaction is as follows:
H3BO3 + H2O ⇌ B(OH)4- + H3O+
We can represent this reaction in terms of the equilibrium constant, Ka, which is given as 5.8 x 10^(-10):
Ka = [B(OH)4-] x [H3O+] / [H3BO3]
Given that the initial concentration of boric acid is 1.0 mol L^-1 and assuming that the equilibrium concentration of [B(OH)4-] is x, [H3O+] is x, and [H3BO3] is (1.0 - x), we can write the expression for Ka as:
5.8 x 10^(-10) = x * x / (1.0 - x)
Now, we can solve this quadratic equation for x:
5.8 x 10^(-10) = x^2 / (1.0 - x)
Rearranging the equation:
5.8 x 10^(-10) * (1.0 - x) = x^2
5.8 x 10^(-10) - 5.8 x 10^(-10) * x = x^2
x^2 + 5.8 x 10^(-10) * x - 5.8 x 10^(-10) = 0
We can now solve this equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Here, a = 1, b = 5.8 x 10^(-10), and c = -5.8 x 10^(-10).
Calculating the discriminant, b^2 - 4ac:
Discriminant = (5.8 x 10^(-10))^2 - 4(1)(-5.8 x 10^(-10))
Discriminant = 3.364 x 10^(-19) + 2.32 x 10^(-9)
Discriminant = 2.656 x 10^(-9)
Since the discriminant is positive, there are two real solutions for x.
Using the quadratic formula:
x1 = (-5.8 x 10^(-10) + √(2.656 x 10^(-9))) / 2(1)
x2 = (-5.8 x 10^(-10) - √(2.656 x 10^(-9))) / 2(1)
Calculating x1 and x2:
x1 ≈ 1.00 x 10^(-5)
x2 ≈ -5.81 x 10^(-10)
Since the concentration of [H3O+] cannot be negative, we disregard x2.
Now, we can calculate the pH of the solution using the equation:
pH = -log[H3O+]
Taking the negative logarithm of x1:
pH ≈ -log(1.00 x 10^(-5))
pH ≈ 5.0
Therefore, the pH of a 1.0 mol L^-1 boric acid solution is approximately 5.0.
To calculate the pH of a solution, you need to know the concentration of the hydronium ion (H3O+). In the case of boric acid (H3BO3), we need to first determine the concentration of H3O+ from the dissociation of the acid.
The dissociation of boric acid can be represented as follows:
H3BO3 ⇌ H+ + H2BO3-
The equilibrium constant for this reaction is given by Ka = [H+][H2BO3-] / [H3BO3]. Given the value of Ka = 5.8 × 10^-10, we can assume that the dissociation of boric acid is minimal, as it is a weak acid.
Since boric acid is a monoprotic acid, the concentration of H+ formed from the dissociation of H3BO3 is equal to the concentration of H3BO3 initially. So, the concentration of H+ is 1.0 mol L^-1.
Now, to calculate the pH, we can use the equation:
pH = -log[H+]
Substituting the value of [H+] = 1.0 mol L^-1 into the equation, we get:
pH = -log(1.0)
= -0
= 0
Therefore, the pH of a 1.0 mol L^-1 boric acid solution is 0.
The molar mass is not required to calculate the pH in this case since the concentration of the acid is provided directly. The -log function is used to calculate the pH, and the equilibrium constant (Ka) is used to determine the concentration of the hydronium ion (H+) in the solution.