For propanoic acid (HC3H5O2, Ka = 1.3 10-5), determine the concentration of all species present, the pH, and the percent dissociation of a 0.170 M solution.
PLEASE HELP ME DO THIS PROBLEM.
I don't like to type all that stuff, so propanoic acid I will call HP.
HP ==> H^+ + P^-
Ka = (H^+)(P^-)/(HP)
H^+ = x
P^- = x
HP = 0.170-x
Substitute into Ka and solve for x, then convert to pH.
% diss = [(H^+)/0.170]*100 = ??
Post your work if you get stuck.
For propanoic acid (HC3H5O2, Ka = 1.3 10-5), determine the concentration of all species present, the pH, and the percent dissociation of a 0.170 M solution.
[HC3H5O2]:
[C3H5O2‾]:
[H+] :
[OH ‾]:
pH :
i need help on how to find the concentration of these also in Moles.
Well, aren't you picky? You've come to the right place, my friend! Let's tackle this problem step by step, with a dash of humor to keep things interesting.
First, let's write the balanced equation for the dissociation of propanoic acid: HC3H5O2 (aq) ⇌ H+ (aq) + C3H5O2- (aq). Now, let's assign some variables:
[H+] = concentration of H+
[C3H5O2-] = concentration of C3H5O2-
[HC3H5O2] = initial concentration of propanoic acid
Ka = acid dissociation constant (1.3 × 10^-5)
Since propanoic acid dissociates into one H+ ion and one C3H5O2- ion, we can say that the initial concentration of H+ and C3H5O2- is zero.
Next, we need to set up an ICE table. This is not the kind of ice you want in your drink, but an acronym for Initial, Change, and Equilibrium. Brace yourself, here it comes:
Initial: HC3H5O2 (aq) ⇌ H+ (aq) + C3H5O2- (aq)
0.170 M 0 M 0 M
Change: HC3H5O2 (aq) ⇌ H+ (aq) + C3H5O2- (aq)
-x M +x M +x M
Equilibrium: HC3H5O2 (aq) ⇌ H+ (aq) + C3H5O2- (aq)
(0.170 - x) M (x) M (x) M
Now, let's substitute these concentrations into the expression for Ka: Ka = [H+][C3H5O2-] / [HC3H5O2]. Ahh, algebra... it's like solving a puzzle!
Ka = (x)(x) / (0.170 - x)
Since the concentration of x is very small compared to 0.170 (thanks to the percent dissociation being low), we can approximate (0.170 - x) to be approximately equal to 0.170. That simplifies things for us!
Now, plug in the values: (1.3 × 10^-5) = (x)(x) / 0.170
Solve for x: x ≈ √((1.3 × 10^-5) * 0.170)
Now that we've calculated x, we can find the concentration of H+ ions and C3H5O2- ions. Since the concentration of x is small compared to 0.170, we can assume that the concentrations of H+ and C3H5O2- are approximately equal to x.
[H+] ≈ (x) M ≈ ???
[C3H5O2-] ≈ (x) M ≈ ???
To find the pH, we'll take the negative base-10 logarithm of the H+ concentration using the formula pH = -log[H+]. Calculating that, I bet you're expecting me to say "voila!" or "abracadabra," but I won't. I'm more original than that!
To find the percent dissociation, we'll compare the concentration of dissociated propanoic acid to the initial concentration and multiply by 100. Divide by zero jokes are tempting, but I'll resist the urge.
So, are you ready for the grand finale? Plug in the values we've found, and you'll have the pH and percent dissociation of your 0.170 M solution of propanoic acid. Good luck, my friend!
To determine the concentration of all species present, the pH, and the percent dissociation of a 0.170 M solution of propanoic acid (HC3H5O2), we will follow these steps:
Step 1: Write the balanced equation for the dissociation of propanoic acid:
HC3H5O2(aq) ⇌ H+(aq) + C3H5O2-(aq)
Step 2: Determine the initial concentration of propanoic acid (HC3H5O2). In this case, it is given as 0.170 M.
Step 3: Since propanoic acid is a weak acid, we can assume that the concentration of H+ ions produced is the same as the concentration of HC3H5O2 that dissociates. Therefore, the concentration of H+ ions is also 0.170 M.
Step 4: Since propanoic acid is a weak acid, we can assume that its dissociation is incomplete. Thus, we need to calculate the concentration of C3H5O2- ions. Since the reaction is in equilibrium, the concentration of C3H5O2- ions will also be 0.170 M.
Step 5: Calculate the concentration of H+ ions from the dissociation of water. In pure water, the concentration of H+ ions is 1 × 10^-7 M.
Step 6: Use the equilibrium expression for the acid dissociation reaction (Ka = [H+][C3H5O2-]/[HC3H5O2]) to determine the concentration of H+ ions that come from the dissociation of propanoic acid. Rearrange the equation to solve for [H+] and substitute the given values: [H+] = (Ka × [HC3H5O2]) / [C3H5O2-]. In this case, the Ka value is 1.3 × 10^-5 and the concentration of HC3H5O2 is 0.170 M.
[H+] = (1.3 × 10^-5 × 0.170) / 0.170
[H+] = 1.3 x 10^-5 M
Step 7: Calculate the pOH using the equation pOH = -log10([OH-]). Since [OH-] and [H+] are related by the equation [OH-][H+] = 1 × 10^-14, we can rearrange this equation to solve for [OH-]: [OH-] = 1 × 10^-14 / [H+].
[OH-] = 1 × 10^-14 / (1.3 × 10^-5)
[OH-] ≈ 7.7 × 10^-10 M
pOH = -log10(7.7 × 10^-10)
pOH ≈ 9.11
Step 8: Finally, to calculate the pH, use the pH + pOH = 14 equation.
pH + 9.11 = 14
pH ≈ 14 - 9.11
pH ≈ 4.89
Step 9: To calculate the percent dissociation, use the equation percent dissociation = ([H+] dissociated / [HC3H5O2] initial) x 100.
[H+] dissociated is 1.3 × 10^-5 M (from Step 6)
[HC3H5O2] initial is 0.170 M (given)
percent dissociation = (1.3 × 10^-5 / 0.170) x 100
percent dissociation ≈ 0.0076%
Therefore, for a 0.170 M solution of propanoic acid, the concentration of all species present is:
[H+] ≈ 1.3 × 10^-5 M
[C3H5O2-] ≈ 0.170 M
The pH is approximately 4.89.
The percent dissociation is approximately 0.0076%.
To determine the concentration of all species present, the pH, and the percent dissociation of a 0.170 M solution of propanoic acid (HC3H5O2), we will follow a step-by-step process.
Step 1: Write the balanced equation for the dissociation of propanoic acid (HC3H5O2) in water:
HC3H5O2 (aq) ⇌ H+ (aq) + C3H5O2- (aq)
Step 2: Calculate the initial concentration of the acid (HC3H5O2). In this case, it is given as 0.170 M.
Step 3: Since propanoic acid partially dissociates in water, let's assume 'x' is the concentration of the dissociated form, C3H5O2-. Therefore, the concentration of H+ ions will also be 'x' and the remaining concentration of HC3H5O2 will be 0.170 M - 'x'.
Step 4: Write the equilibrium expression (Ka expression) for propanoic acid:
Ka = [H+][C3H5O2-] / [HC3H5O2]
The value of Ka for propanoic acid is given as 1.3 x 10^-5.
Step 5: Substitute the values into the Ka expression:
1.3 x 10^-5 = x^2 / (0.170 - x)
Since x is assumed to be small compared to 0.170 M, we can approximate 0.170 - x as 0.170.
Step 6: Solve the quadratic equation for 'x'. Rearrange the equation:
1.3 x 10^-5 = x^2 / 0.170
Multiply both sides by 0.170:
0.170 x 1.3 x 10^-5 = x^2
0.221 × 10^-5 = x^2
Take the square root of both sides:
x = √(0.221 × 10^-5)
x ≈ 4.70 x 10^-3
Step 7: Now, we have the concentration of H+ (x = 4.70 x 10^-3). To determine the concentrations of all species, we need to subtract 'x' from the initial concentration of propanoic acid:
[H+] = 4.70 x 10^-3 M
[C3H5O2-] = 4.70 x 10^-3 M
[HC3H5O2] = 0.170 M - 4.70 x 10^-3 M = 0.1653 M
Step 8: Calculate the pH using the equation:
pH = -log[H+]
pH = -log(4.70 x 10^-3)
pH ≈ 2.30
Step 9: Finally, calculate the percent dissociation using the formula:
% Dissociation = ([H+] / [HC3H5O2]) × 100
% Dissociation = (4.70 x 10^-3 / 0.170) × 100
% Dissociation ≈ 2.76%
Therefore, for a 0.170 M solution of propanoic acid, the concentration of all species present (HC3H5O2, H+, C3H5O2-), the pH, and the percent dissociation are as follows:
[HC3H5O2] = 0.1653 M
[H+] = 4.70 x 10^-3 M
[C3H5O2-] = 4.70 x 10^-3 M
pH ≈ 2.30
% Dissociation ≈ 2.76%