A 30.0-ml sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. What is the pH at 5 ml of added base, at 10 ml of added base?
Use the henderson-hasselbalch equation
pH=pKa+log[A-/HA]
where
Pka=-logKa
Ka=1.3 x 10–5
HA=propanoic acid
A-=conjugate base of propanic acid
0.0300L *(0.165M)=moles of propanoic acid
0.010*(0.300M)= moles of KOH
0.005*(0.300M)= moles of KOH
Solve for moles of conjugate base of propanoic acid.
moles of propanoic acid-moles of KOH=moles of conjugate base of propanoic acid
moles of propanoic acid-moles of conjugate base of propanoic acid=moles of propanoic acid
***For the first problem, total volume is 40mL or 0.040 L
moles of propanoic acid/total volume=HA
moles of conjugate base of propanoic acid/total volume=A-
Substitute into the henderson hasselbalch equation and solve.
To find the pH at different points during the titration, you will need to determine the concentration of the propanoic acid and its conjugate base (propanoate) at each point. From there, you can use the Henderson-Hasselbalch equation to calculate the pH.
First, let's calculate the initial concentration of propanoic acid in the 30.0 ml sample. Since the concentration is given as 0.165 M, we can directly use this value.
Next, let's calculate the moles of propanoic acid present in the 30.0 ml sample:
moles of propanoic acid = concentration * volume
moles of propanoic acid = 0.165 M * 0.030 L
moles of propanoic acid = 0.00495 moles
At this point, no base has been added yet, so the concentration of propanoic acid remains the same.
Now, let's consider the addition of 5 ml of 0.300 M KOH. Since KOH is a strong base, it will react completely with the propanoic acid to form the propanoate ion (C3H5O2-). The reaction equation is:
C3H5COOH (propanoic acid) + KOH → C3H5COOK (propanoate) + H2O
Since the reaction is a 1:1 ratio, the moles of propanoic acid reacted will be the same as the moles of KOH added.
moles of propanoic acid reacted = 0.00495 moles
Now, we need to calculate the volume of the solution after adding 5 ml of base:
total volume = initial volume + volume of base added
total volume = 30.0 ml + 5.0 ml
total volume = 35.0 ml
Next, calculate the concentration of the propanoate ion using the moles and volume:
concentration of propanoate = moles of propanoate / total volume
concentration of propanoate = 0.00495 moles / 0.035 L
concentration of propanoate = 0.141 M
Now that we have the concentrations of both propanoic acid (0.165 M) and propanoate (0.141 M), we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log ([A-] / [HA])
The pKa value for propanoic acid is typically around 4.87.
For 5 ml of added base:
pH = 4.87 + log(0.141 / 0.165)
To find the pH at 10 ml of added base, repeat the calculations using the new volume (40.0 ml) and the updated concentration of propanoate based on the amount reacted.
I hope this explanation helps you understand how to calculate the pH at different points during the titration!
To find the pH at different volumes of added base, we need to calculate the moles of propanoic acid and KOH at each point and use the Henderson-Hasselbalch equation to calculate the pH.
Step 1: Calculate the moles of propanoic acid initially present:
Moles of propanoic acid = concentration × volume
Moles of propanoic acid = 0.165 M × 30.0 ml = 4.95 mmol
Step 2: Calculate the volume of added base:
At 5 ml of added base, the total volume will be 30.0 ml (initial volume) + 5 ml (added base volume) = 35 ml.
At 10 ml of added base, the total volume will be 30.0 ml (initial volume) + 10 ml (added base volume) = 40 ml.
Step 3: Calculate the moles of KOH:
Moles of KOH = concentration × volume
At 5 ml of added base: Moles of KOH = 0.300 M × 5 ml = 1.50 mmol
At 10 ml of added base: Moles of KOH = 0.300 M × 10 ml = 3.00 mmol
Step 4: Calculate the moles of propanoic acid remaining at each point:
At 5 ml of added base: Moles of propanoic acid remaining = Initial moles - moles of KOH reacted
Moles of propanoic acid remaining = 4.95 mmol - 1.50 mmol = 3.45 mmol
At 10 ml of added base: Moles of propanoic acid remaining = Initial moles - moles of KOH reacted
Moles of propanoic acid remaining = 4.95 mmol - 3.00 mmol = 1.95 mmol
Step 5: Calculate the moles of propanoate formed at each point:
At 5 ml of added base: Moles of propanoate formed = moles of KOH reacted
Moles of propanoate formed = 1.50 mmol
At 10 ml of added base: Moles of propanoate formed = moles of KOH reacted
Moles of propanoate formed = 3.00 mmol
Step 6: Calculate the concentrations of propanoic acid and propanoate at each point:
Concentration of propanoic acid = Moles of propanoic acid remaining / Total volume
At 5 ml of added base: Concentration of propanoic acid = 3.45 mmol / 35 ml = 0.0986 M
At 10 ml of added base: Concentration of propanoic acid = 1.95 mmol / 40 ml = 0.0488 M
Concentration of propanoate = Moles of propanoate formed / Total volume
At 5 ml of added base: Concentration of propanoate = 1.50 mmol / 35 ml = 0.0429 M
At 10 ml of added base: Concentration of propanoate = 3.00 mmol / 40 ml = 0.0750 M
Step 7: Calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
where pKa is the acid dissociation constant and [A-] and [HA] are the concentrations of the conjugate base and acid, respectively.
The pKa value for propanoic acid is 4.87.
At 5 ml of added base:
pH = 4.87 + log (0.0429 / 0.0986)
At 10 ml of added base:
pH = 4.87 + log (0.0750 / 0.0488)
Calculating the final pH values will give the pH at 5 ml and 10 ml of added base.