What is the pH of a solution by dissolving 8.00 mmol of NaOAC in 200 mL of 0.100 m HOAC. (you need the pka for acetic acid)
A word to the wise. Do NOT (that's NOT) think 0.1 m HOAc is the same as 0.1 M HOAc. 0.1 m is 0.1 MOLAL while 0.1 M is 0.1 MOLAR. They are NOT the same.
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
base = NaOAc; acid = HOAc.
To determine the pH of the solution, we will use the Henderson-Hasselbalch equation. The equation is given by:
pH = pKa + log [A-] / [HA]
First, we need the pKa value for acetic acid. The pKa of acetic acid (HOAC) is approximately 4.75.
Next, we need to determine the concentrations of the acetate ion (A-) and acetic acid (HA) in the solution.
Given:
- 8.00 mmol of NaOAC (sodium acetate) is dissolved.
- Volume of the solution = 200 mL (0.200 L)
- Concentration of HOAC = 0.100 M
To find the concentration of acetate ion (A-):
- Convert the mmol of NaOAC to moles: 8.00 mmol = 0.008 moles
- Since NaOAC dissociates into Na+ and OAC-, the concentration of acetate ion (A-) is also 0.008 moles.
To find the concentration of acetic acid (HA):
- Concentration = Molarity x Volume
- Concentration = 0.100 M x 0.200 L = 0.020 moles
Now we can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log [A-] / [HA]
pH = 4.75 + log (0.008 / 0.020)
Calculating the logarithm:
Log (0.008 / 0.020) = -0.60
Now, we can calculate the pH:
pH = 4.75 + (-0.60)
pH = 4.15
Therefore, the pH of the solution is approximately 4.15.
To find the pH of a solution after dissolving a certain amount of a weak acid and its conjugate base, we need to consider the dissociation of the weak acid. In this case, we have a solution containing sodium acetate (NaOAC) and acetic acid (HOAC). The pH of the solution will depend on the concentration of the weak acid (acetic acid) and its dissociation constant (Ka or pKa).
First, let's find the concentration of acetic acid (HOAC) in the 200 mL solution. We're given that the concentration of HOAC is 0.100 M. To convert it to moles (mmol), we can use the formula:
Concentration (M) = moles (mmol) / volume (L)
Rearranging the formula, we can find the number of moles:
moles (mmol) = concentration (M) x volume (L)
Given that the volume is 200 mL, which is equal to 0.200 L:
moles (mmol) = 0.100 M x 0.200 L = 0.020 mmol
So, we have 0.020 mmol of acetic acid (HOAC).
Next, we need to determine the concentration of sodium acetate (NaOAC) in the solution. Since NaOAC dissociates completely in water, the concentration of NaOAC is equal to the concentration of the sodium ion, which is determined by the moles of NaOAC and the volume of the solution. In this case, we are given that we have 8.00 mmol of NaOAC.
moles of NaOAC = 8.00 mmol
Now, let's calculate the moles of sodium ion (Na+):
moles of Sodium ion (Na+) = moles of NaOAC = 8.00 mmol
To calculate the final concentration of acetic acid (HOAC) and sodium ion (Na+), we need to consider the total volume of the solution. The volume is given as 200 mL, which is equal to 0.200 L.
Concentration (M) = moles (mmol) / volume (L)
For acetic acid (HOAC), we have:
Concentration of HOAC = 0.020 mmol / 0.200 L = 0.100 M
And for sodium ion (Na+), using the given moles:
Concentration of sodium ion (Na+) = 8.00 mmol / 0.200 L = 40.0 M
Now, we can use the equilibrium equation for acetic acid dissociation to find the pH of the solution. The equation is as follows:
HOAC(aq) + H2O(l) ⇌ H3O+(aq) + OAC-(aq)
The pKa for acetic acid (HOAC) is given in the problem statement. For acetic acid, the pKa is typically around 4.76.
To find the pH, we can use the Henderson-Hasselbalch equation, which relates the pH and the concentrations of the acid (HOAC) and its conjugate base (OAC-):
pH = pKa + log([OAC-] / [HOAC])
Plugging in the known values, we get:
pH = 4.76 + log(40.0 / 0.100)
At this point, you can use a scientific calculator or software to evaluate the logarithm and calculate the pH. The final pH will depend on the actual values obtained from the calculations.
Remember to consider significant figures throughout the calculations and to apply appropriate rounding at the end.