(a) Find pH and pOH of 1.0 M solution of sodium benzoate, NaC6H5COO. (Ka of benzoic acid is 6.2x10^-5)
(b) Calculate pH after 0.205 moles per liter of HCl was added (assume volume does not change
(c) Calculate pH after 1.0 moles per liter of HCl was added (assume volume does not change)
So for part (a) I know how to start it:
We have
NaC6H5COO (aq) <---> C6H5COO- (aq) + Na+ (aq)
C6H5COO- will react with water:
C6H5COO- (aq) + H2O (l) <---> C6H5COOH (aq) + Na+ (aq)
initial M 1.0 <---> 0 + 0
change -x <---> +x , +x
equilib 1.0-x <---> x, x
Since we're solving for the base C6H5COO-
Kb = [C6H5COOH][OH-] / [C6H5COO-]
But I have the Ka, not Kb. Would I just plug Ka in for C6H5COOH?
For part (b)
NaC6H5COO- (aq) + HCl (aq) <---> NaCl (aq) + C6H5COOH (aq)
I'm left with 0.8 NaC6H5COO-, 0.205 NaCl, 0.205 C6H5COOH so I can use Henderson-Hasselbach equation
pH = pKa + log (0.8/0.205)
Am I going to use the pKa of the Ka of benzoic acid given?
NaC6H5COO- (aq) + HCl (aq) <---> NaCl (aq) + C6H5COOH (aq)
should say
NaC6H5COO (aq) + HCl (aq) <---> NaCl (aq) + C6H5COOH (aq)
and 0.8 NaC6H5COO, not 0.8 NaC6H5COO-
Just the a part here. Yes, you have Ka, not Kb but you can calculate Kb BECAUSE Ka*Kb = Kw OR Kb = Kw/Ka.
For part b here.
Yes, you use pKa for benzoic acid.
One of the other values I have to plug in for HCl is 0.803 mols/L
So then at equilibrium, NaC6H5COO is 0.197, NaCl is 0.803, C6H5COOH is 0.803
If I plug it into Henderson Hasselbach equation:
pH = pKa + log [0.197]/[0.803]
the log of 0.197/0.803 is negative, -0.61
Can I still plug it even if it's negative?
pH = 4.21 + (-0.61) = 3.60
Conceptually it makes sense because weak base-strong acid titrations are less than 7, and adding more HCl will decrease the pH. I just want to make sure that I can use negative numbers using the equation.
Sure. You aren't substituting a negative number. You can't do that. You are calculating a value of log (0.197/0.803) and adding that to the pKa of the benzoic acid. The log term will be positive in some situations and negative in others. That is a mathematical operation. [By the way, I've seen the HH equation written as
pH = pKa - log(acid)/(base) but most of the time its a + ratio and that makes the ratio be reversed.]And of course that's why the HH equation works so nicely because adding more acid should make the solution more acidic.
R-NH2 + HCl
For part (a), you're on the right track. Since sodium benzoate, NaC6H5COO, is the conjugate base of benzoic acid, C6H5COOH, we can assume that the reaction between C6H5COO- and H2O will give us the corresponding conjugate acid, C6H5COOH.
The equation for the reaction is:
C6H5COO- (aq) + H2O (l) ⇌ C6H5COOH (aq) + OH- (aq)
To find the pH and pOH of the solution, we need to determine the concentration of OH- ions in the solution, which is formed from the dissociation of water as well as the reaction mentioned above. We can use the equilibrium expression for Kb:
Kb = [C6H5COOH][OH-] / [C6H5COO-]
However, since we have the Ka value for benzoic acid instead of the Kb value, we can use the relationship between Ka and Kb for a conjugate acid-base pair. The relationship is:
Ka * Kb = Kw
Where Kw is the equilibrium constant for water, which is 1.0x10^-14 at 25°C. So, rearranging the equation, we get:
Kb = Kw / Ka
Now you can substitute the given Ka value into the equation to find the Kb value. Once you have the Kb value, you can solve for the concentration of OH- ions in the solution and then calculate the pOH using the equation pOH = -log [OH-]. Finally, you can find the pH using the equation pH = 14 - pOH.
For part (b), you correctly identified that you can use the Henderson-Hasselbalch equation in this case, since you have the concentration of the conjugate acid (C6H5COOH) and the conjugate base (C6H5COO-). The Henderson-Hasselbalch equation is:
pH = pKa + log ([A-] / [HA])
In this case, [A-] is 0.8 M (0.1 M - 0.205 M) and [HA] is 0.205 M (initial concentration of C6H5COOH). You can use the pKa value provided for benzoic acid to substitute into the equation and find the pH.
For part (c), the process is the same as part (b), except the values for [A-] and [HA] will change based on the additional concentration of HCl added. Using the Henderson-Hasselbalch equation, you can calculate the pH with the new values of [A-] and [HA].