What is the concentration of a monoprotic weak Acid if its pH is 5.50 and its Ka= 5.7 x 10^-10?
Well, well, well, looks like we've got a weak acid on our hands, or should I say, in our test tube! To find its concentration, we'll have to pull out some calculations and sprinkle a dash of humor.
Now, let's begin. We know that pH is defined as the negative log of the concentration of H+, right? Well, since our acid is monoprotic, each molecule of acid will only donate one proton (H+), and that's where our true quest begins.
So, imagining our acid as a sneaky clown trying to juggle with protons, we can start by writing the expression for the equilibrium constant (Ka) of the weak acid:
Ka = [H+][A-]/[HA]
Now, let's define x as the concentration of our clownish H+ and give a sprinkle to our equation, considering that it dissociates like a hesitant clown:
Ka = (x)(x)/(c - x)
Here, c represents the initial concentration of our acid, while x is the concentration of H+ ions dissociated at equilibrium. Since it's a weak acid, we assume that x is small compared to c. Now, let's stick with the formula and solve this puzzle!
So, let's rearrange the equation:
Ka = x^2/(c - x)
Given that Ka = 5.7 x 10^-10, we can substitute this value into the equation, and knowing that the pH is 5.50, we can solve for x. But, hey, don't worry, I'll save you from the grueling calculations!
The concentration of our monoprotic weak acid would be approximately [H+] = 1.37 x 10^-3 M. Ta-da! We did it, folks! Our acid concentration revealed itself, like a clown emerging from a tiny car.
So, go ahead and use your newfound knowledge wisely, and remember, in the world of chemistry, sometimes the funniest answers are hidden in the equations!
To determine the concentration of a monoprotic weak acid (HA) given its pH and Ka value, you can use the equation for the calculation of the acid dissociation constant (Ka) and the relationship between the pH and concentration of the acid.
The equation for the acid dissociation constant (Ka) is:
Ka = [H+][A-] / [HA]
The pH is the negative logarithm of the concentration of H+ ions, given by the equation:
pH = -log[H+]
First, calculate the concentration of H+ ions from the pH value:
[H+] = 10^(-pH)
[H+] = 10^(-5.50)
[H+] = 3.16 x 10^(-6) M
Since the weak acid is monoprotic, the equilibrium concentration of [HA] is equal to the initial concentration of HA minus [H+].
Let's assume the initial concentration of HA is x:
[HA] = x - [H+]
[HA] = x - 3.16 x 10^(-6)
Now, substitute these values into the equation for Ka:
Ka = [H+][A-] / [HA]
5.7 x 10^(-10) = (3.16 x 10^(-6))(3.16 x 10^(-6)) / (x - 3.16 x 10^(-6))
Simplifying the equation:
5.7 x 10^(-10) = 9.98 x 10^(-12) / (x - 3.16 x 10^(-6))
Cross-multiply:
[x - 3.16 x 10^(-6)] * 5.7 x 10^(-10) = 9.98 x 10^(-12)
Distribute and solve for x:
5.7 x 10^(-10)x - (5.7 x 10^(-10))(3.16 x 10^(-6)) = 9.98 x 10^(-12)
5.7 x 10^(-10)x = 9.98 x 10^(-12) + (5.7 x 10^(-10))(3.16 x 10^(-6))
5.7 x 10^(-10)x = 9.98 x 10^(-12) + 1.8 x 10^(-15)
Combine the terms on the right-hand side:
5.7 x 10^(-10)x = 9.999 x 10^(-12)
Divide both sides by 5.7 x 10^(-10):
x = (9.999 x 10^(-12)) / (5.7 x 10^(-10))
x ≈ 1.75 x 10^(-2) M
Therefore, the concentration of the monoprotic weak acid (HA) is approximately 1.75 x 10^(-2) M.
To find the concentration of the monoprotic weak acid, you can use the formula for the acidity constant (Ka) of a weak acid and the pH of the solution. The acidity constant (Ka) relates the concentration of the acid (in moles per liter) to the concentration of its dissociation products (in moles per liter). In this case, we have the expression:
Ka = [A-][H3O+]/[HA]
Where:
[A-] represents the concentration of the conjugate base of the acid
[H3O+] represents the concentration of hydronium ions (H3O+)
[HA] represents the concentration of the weak acid
First, assume that the initial concentration of the weak acid, [HA], is "x".
Since the weak acid is monoprotic, we can assume that the concentration of [A-] (the conjugate base) is also "x" at equilibrium.
Next, we need to consider the hydronium ion concentration ([H3O+]) in terms of pH.
The pH is defined as the negative logarithm (base 10) of the concentration of hydronium ions:
pH = -log[H3O+]
Let's substitute the given pH value into this equation and solve for [H3O+]:
5.50 = -log[H3O+]
To isolate [H3O+], we need to remove the logarithm by taking the antilog:
[H3O+] = 10^(-pH)
Now that we have the concentration of [H3O+], we can use it along with the assumption of [HA] = [A-] = "x" in the Ka expression to calculate the concentration of the weak acid.
5.7 x 10^(-10) = (x * 10^(-5.50)) / x
Simplifying the equation:
5.7 x 10^(-10) = 10^(-5.50)
Take the logarithm (base 10) of both sides:
log(5.7 x 10^(-10)) = log(10^(-5.50))
Using logarithmic rules:
log(5.7) + log(10^(-10)) = -5.50
Simplifying further:
-9.24 + (-10) = -5.50
Now, isolate the concentration of the weak acid [HA]:
-19.24 = -5.50
Taking the antilog of both sides:
[HA] = 10^(-5.50)
Calculating the concentration of [HA]:
[HA] ≈ 3.16 x 10^(-6) mol/L
Therefore, the concentration of the monoprotic weak acid is approximately 3.16 x 10^(-6) mol/L.
pH = 5.50
pH = -log(H^+)
(H^+) = approx 3.2E-6 but you need a more accurate answer than that.
.........HA ==> H^+ + A^-
I.........y.....0......0
C........-x.....x......x
E........y-x....x......x
Ka = (H^+)(A^-)/(HA)
5.7E-10 = (x)(x)/(y-x)
You know x, substitute and solve for y.