1.With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4.
You add 8g sodium hydroxide to this solution, what is the new pH?
2. With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4.
You add 16g hydrobromic acid to this solution, what is the new pH?
3. With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4.
You add 60.7g hydrobromic acid to this solution, what is the new pH?
Assume a 1L solution.
HCOOH is formic acid = 0.85 moles in 1 L = 0.85 M
HCOO^- is formate ion. = 0.75 moles in 1 L = 0.75 M
mols NaOH = grams/molar mass = 8/40 = 0.20 mol in 1 L = 0.20 M
The equation is
...................HCOOH + OH^- ==> HCOO^- + H2O
I....................0.85...........0...............0.75............0
add...............................0.20........................................
C..................-0.20........-0.20............+0.20.................
E..................0.65..............0..............0.95.....................
Plug the E line into the pH = pKa + log [(base)/(acid)] and solve for pH.
The others follow the same procedure except that when you're adding and acid to the buffer the equation is HCOO^- + H^+ ==> HCOOH and the ICE chart changes. Post your work if you get stuck.
Could you help with the second question. I tried making this ice chart but it was wrong
16gHBr/80.9119g/mol = 0.1977 mol in 1L
................HCOO^- + H^+ ==> HCOOH
I................0.85.............0................0.75
C..........-0.1977....-0.1977.......+0.1977
E............0.6523....-0.1977......0.5523
I used that and got 3.82 but it was wrong.
To find the new pH in each scenario, we need to consider the reaction that occurs when the acid or base is added to the buffer solution.
1. When 8g sodium hydroxide (NaOH) is added:
The reaction that occurs is:
NaOH + HCOOH --> HCOONa + H2O
To calculate the new pH, we need to determine the concentration of the resulting formate ion (HCOO-) and formic acid (HCOOH) in the solution.
To find the concentration of formate ion (HCOO-):
8g NaOH = (8g / 40.00g/mol) = 0.20 moles NaOH
Since the mole ratio of NaOH to HCOO- is 1:1, the concentration of HCOO- is also 0.20 moles.
The concentration of the formic acid (HCOOH) remains the same at 0.85 moles.
Now, we can use the Henderson-Hasselbalch equation to find the new pH:
pH = pKa + log ([HCOO-] / [HCOOH])
The pKa value can be found using the Ka value:
pKa = -log(Ka) = -log(1.8x10^-4) = 3.74
Substituting the values:
pH = 3.74 + log (0.20 / 0.85) = 3.74 - 0.79 = 2.95
Therefore, the new pH is approximately 2.95.
2. When 16g hydrobromic acid (HBr) is added:
The reaction that occurs is:
HBr + HCOONa --> NaBr + HCOOH
To calculate the new pH, we need to determine the concentration of the resulting formate ion (HCOO-) and formic acid (HCOOH) in the solution.
To find the concentration of formate ion (HCOO-):
16g HBr = (16g / 80.91g/mol) = 0.20 moles HBr
Since the mole ratio of HBr to HCOO- is 1:1, the concentration of HCOO- is also 0.20 moles.
The concentration of the formic acid (HCOOH) remains the same at 0.85 moles.
Using the Henderson-Hasselbalch equation, we can find the new pH.
pKa = -log(Ka) = -log(1.8x10^-4) = 3.74
Substituting the values:
pH = 3.74 + log (0.20 / 0.85) = 3.74 - 0.79 = 2.95
Therefore, the new pH is approximately 2.95.
3. When 60.7g hydrobromic acid is added:
The reaction that occurs is the same as in the previous scenario:
HBr + HCOONa --> NaBr + HCOOH
To calculate the new pH, we need to determine the concentration of the resulting formate ion (HCOO-) and formic acid (HCOOH) in the solution.
To find the concentration of formate ion (HCOO-):
60.7g HBr = (60.7g / 80.91g/mol) = 0.75 moles HBr
Since the mole ratio of HBr to HCOO- is 1:1, the concentration of HCOO- is also 0.75 moles.
The concentration of the formic acid (HCOOH) remains the same at 0.85 moles.
Using the Henderson-Hasselbalch equation, we can find the new pH.
pKa = -log(Ka) = -log(1.8x10^-4) = 3.74
Substituting the values:
pH = 3.74 + log (0.75 / 0.85) = 3.74 - 0.094 = 3.646
Therefore, the new pH is approximately 3.65.