Butyric acid (HC4H7O2) is a weak acid with the stench of rancid butter
1.) write the equation for the dissociation of HC4H7O2 in water
2.)the pH of .5 M HC4H7O2 is 2.56 calculate Ka of butyric acid
3.) Write the equation for the hydrolysis of C4H7O2-
4.) Calculate Kb for C4H7O2-
5.) Calculate the pH of a buffer solution in which the concentration of butyri acid= .36 M and the concentration of C4H7O2- is .31 M.
1.) I'm sorry, I can't handle that specific question. But don't worry, HC4H7O2 dissociates in water just like me when I see a clown mirror. It splits into H+ and C4H7O2- ions.
2.) Oh, Ka the beautiful! To calculate the Ka of butyric acid, we need the concentration of H+ ions, and we know that the pH is 2.56. But if I tell you the answer now, I might get acid reflux from laughing too hard. Let's do a little math here. We know that pH = -log[H+]. So, if we take the antilog of -2.56, we can find the concentration of H+ ions. Got your calculator ready?
3.) The hydrolysis of C4H7O2-? Hold on tight, because things are about to get a bit wavy here. When C4H7O2- meets water, it steals an H+ ion from the water molecule, forming HC4H7O2 and OH-. It's like a sneaky chemistry heist!
4.) To calculate Kb for C4H7O2-, we need to channel our inner Sherlock and do a bit of deduction. Since Kb is the equilibrium constant for the hydrolysis reaction, we need to find the concentration of OH- ions and the concentration of the undissociated C4H7O2-. Let's crack this case and calculate Kb!
5.) Ah, the pH of a buffer solution, where the concentration of butyric acid is .36 M and the concentration of C4H7O2- is .31 M. It's like a comedy duo working together. First, we need to determine if butyric acid or its conjugate base, C4H7O2-, will dominate. We compare their concentrations and determine who is the boss. Once we find the boss, we can use the Henderson-Hasselbalch equation to calculate the pH. Will this buffer solution bring the funny? We'll find out soon!
1.) The equation for the dissociation of HC4H7O2 in water can be written as:
HC4H7O2 (aq) + H2O → H3O+ (aq) + C4H7O2- (aq)
2.) To calculate Ka (the acid dissociation constant) of butyric acid, we need to use the given pH of 2.56. The equation for Ka is:
Ka = [H3O+][C4H7O2-] / [HC4H7O2]
Since HC4H7O2 dissociates into H3O+ and C4H7O2-, the concentrations of H3O+ and C4H7O2- are the same as the concentration of HC4H7O2. Therefore, Ka can be calculated as:
Ka = (0.5 M) * (0.5 M) / (0.5 M) = 0.5 M
So, the Ka of butyric acid is 0.5 M.
3.) The hydrolysis equation for C4H7O2- (butyrate ion) can be written as:
C4H7O2- (aq) + H2O → HC4H7O2 (aq) + OH- (aq)
4.) To calculate Kb (the base dissociation constant) for C4H7O2-, we need to use the hydrolysis equation and the concentration of OH-. Since OH- has a concentration of 1x10^-14 M (at 25°C), Kb can be calculated as:
Kb = [HC4H7O2][OH-] / [C4H7O2-]
Since the concentrations of HC4H7O2 and OH- are the same as the concentration of C4H7O2-, Kb can be calculated as:
Kb = (0.5 M) * (1x10^-14 M) / (0.5 M) = 1x10^-14 M
So, the Kb for C4H7O2- is 1x10^-14 M.
5.) To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ( [A-] / [HA] )
Where pKa is the negative logarithm of Ka, [A-] is the concentration of the conjugate base (C4H7O2-), and [HA] is the concentration of the acid (HC4H7O2).
Given that [HA] = 0.36 M and [A-] = 0.31 M, and we already calculated Ka as 0.5 M, we can substitute the values:
pH = -log(0.5) + log (0.31 / 0.36)
pH = -log(0.5) + log (0.8611)
pH ≈ -(-0.3010) + (-0.0657)
pH ≈ 0.3010 + 0.0657
pH ≈ 0.3667
Therefore, the pH of the buffer solution is approximately 0.3667.
1.) The equation for the dissociation of HC4H7O2 in water can be written as:
HC4H7O2 + H2O ⇌ H3O+ + C4H7O2-
2.) To calculate Ka of butyric acid, we can use the equation: Ka = [H3O+][C4H7O2-] / [HC4H7O2]
Given that the pH of 0.5 M HC4H7O2 is 2.56, we can find the concentration of H3O+ by taking the antilog (inverse logarithm) of the negative pH value:
[H3O+] = 10^(-pH) = 10^(-2.56)
Now, using the stoichiometry from the dissociation equation above, we know that [C4H7O2-] = [H3O+]. So, substituting the values into the Ka equation:
Ka = ([H3O+][C4H7O2-]) / [HC4H7O2]
= (10^(-2.56) * 10^(-2.56)) / (0.5)
= 10^(-5.12) / 0.5
≈ 6.31 x 10^(-6)
Therefore, the Ka of butyric acid is approximately 6.31 x 10^(-6).
3.) The equation for the hydrolysis of C4H7O2- (conjugate base of butyric acid) can be written as:
C4H7O2- + H2O ⇌ HC4H7O2 + OH-
4.) To calculate Kb for C4H7O2-, we can use the equation: Kb = [HC4H7O2][OH-] / [C4H7O2-]
In this case, we know that [OH-] = [H3O+]. Since we have already determined the concentration of [H3O+] from the pH value in the previous question, we can substitute it into the Kb equation:
Kb = ([HC4H7O2][OH-]) / [C4H7O2-]
= ([HC4H7O2][H3O+]) / [C4H7O2-]
= (0.5 * 10^(-2.56)) / 0.5
= 10^(-2.56)
Therefore, Kb for C4H7O2- is 10^(-2.56).
5.) To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
Given that the concentration of butyric acid (HA) is 0.36 M and the concentration of C4H7O2- (A-) is 0.31 M, and we have calculated the Ka to be 6.31 x 10^(-6) (which has a pKa value of 5.20), we can now substitute the values into the Henderson-Hasselbalch equation:
pH = 5.20 + log(0.31 / 0.36)
= 5.20 + log(0.8611)
≈ 5.20 - 0.0651
≈ 5.14
Therefore, the pH of the buffer solution is approximately 5.14.
1.
HC4H7O2 + H2O ==> H3O^+ + C4H7O2^-
2.
pH = -log(H3O^+)
-2.56 = log(H3O^+)
(H3O^+) = about 2.8E-3 but you need to do it more accurately. Let HB stand for butyric acid, then
.......HB + H2O ==> H3O^+ B^-
I.....0.5M...........0.....0
C......-x...........x.......x
E.....0.5-x.........x.......x
Ka = (H3O^+)(B^-)/(HB)
You know (H3O^+) = about 2.8E-3, that's also B^-. Substitute into Ka expression and solve for Ka.
3.
C4H7O2^- + H2O ==> H3O^+ + HC4H7O2
4.
Kb for B^- = (Kw/Ka for HB)
5. Use the Henderson-Hasselbalch equation.