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-
(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?
6 answers
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.
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.
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.
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
1 answer