A 40 mL sample of 0.25 M NaCHOO is
titrated with 0.2 M HCl. What is the pH of the solution after 140 mL of HCl has been added?
40 mL x 0.25M HCOONa = 10 mmoles.
140 mL x 0.2M HCl = 28 mmoles.
..............HCOONa + HCl ==>HCOOH + NaCl
begin.........10.......0........0
change....... -10.....+28......+10...+10
final..........0......+18.......+10..+10
So your question is asking for the pH of 18 mmoles HCl + 10 mmoles HCOOH in (140mL + 40 mL = 180 mL). A good estimate is to ignore the HCOOH (much weaker than HCL) and you have 18 mmoles/180 mL = 0.1 M = pH of 1.0.
If you want to do it up brown, you can calculate the amount of H^+ contributed by HCOOH (M = mmoles/mL = 10/180 =0.0556M) as follows (remember that the HCl provides a common ion (H^+) and that will shift the HCOOH ionization to the left and DECREASE the production of H^+ from the HCOOH source.):
.............HCOOH ==> H^+ + HCOO^-.
begin.......0.0556.....0.1....0
change........-x.......+x.....+x
final......0.00556-x....x......x
(H^+)(HCOO^-)/(HCOOH) = Ka
Look up Ka in your text or notes and solve for x. I have estimated it at about 1E-6 and that added to 0.1M is inconsequential.
To find the pH of the solution after adding 140 mL of 0.2 M HCl, we need to calculate how many moles of HCl are added and how this affects the concentration of the resulting solution.
Step 1: Calculate the number of moles of HCl added.
The concentration of HCl is 0.2 M, which means there are 0.2 moles of HCl per liter (1000 mL) of solution.
Therefore, in 140 mL of HCl, there are (0.2 moles/L) * (140 mL / 1000 mL/L) = 0.028 moles of HCl.
Step 2: Determine the excess amount of HCl present.
Since the ratio of NaCHOO to HCl is 1:1, the amount of HCl added is in excess compared to NaCHOO.
To find the amount of excess HCl, we subtract the moles of HCl added from the initial moles of NaCHOO.
Initial moles of NaCHOO = concentration * volume = (0.25 moles/L) * (40 mL / 1000 mL/L) = 0.01 moles.
Excess HCl = initial moles of NaCHOO - moles of HCl added = 0.01 moles - 0.028 moles = -0.018 moles.
Step 3: Calculate the final volume of the solution.
The initial volume was 40 mL, and 140 mL of HCl was added. Therefore, the final volume is 40 mL + 140 mL = 180 mL.
Step 4: Calculate the final concentration of the solution.
The concentration is given by the formula:
Final concentration = (Initial moles - Excess moles) / Final volume
Final concentration = (0.01 moles - (-0.018 moles)) / (180 mL / 1000 mL/L)
Final concentration = (0.01 moles + 0.018 moles) / 0.18 L
Final concentration = 0.028 moles / 0.18 L
Final concentration = 0.1556 M
Step 5: Calculate the pH of the solution.
The pH of a solution can be calculated using the formula:
pH = -log[H+]
Since HCl is a strong acid, it dissociates completely, and the concentration of H+ ions will be equal to the final concentration of the solution. Therefore:
pH = -log(0.1556) = 1.81
So, the pH of the solution after adding 140 mL of HCl is 1.81.
To find the pH of the solution after adding 140 mL of HCl, we first need to determine the moles of HCl and NaCHOO in the solution.
Given:
- Initial volume of NaCHOO solution = 40 mL
- Concentration of NaCHOO = 0.25 M
- Volume of HCl added = 140 mL
- Concentration of HCl = 0.2 M
Step 1: Calculate the moles of NaCHOO in the initial solution.
Moles of NaCHOO = concentration × volume
Moles of NaCHOO = 0.25 M × (40 mL / 1000 mL/1 L)
Moles of NaCHOO = 0.01 moles
Step 2: Calculate the moles of HCl added.
Moles of HCl = concentration × volume
Moles of HCl = 0.2 M × (140 mL / 1000 mL/1 L)
Moles of HCl = 0.028 moles
Step 3: Determine the limiting reactant.
Since the moles of HCl (0.028 moles) is higher than the moles of NaCHOO (0.01 moles), HCl is the limiting reactant. Therefore, all of the NaCHOO will react, and some HCl will be left over.
Step 4: Find the excess moles of HCl.
Excess moles of HCl = moles of HCl added - moles of NaCHOO reacted
Excess moles of HCl = 0.028 moles - 0.01 moles
Excess moles of HCl = 0.018 moles
Step 5: Calculate the final volume of the solution.
Final volume of the solution = initial volume of NaCHOO + volume of HCl added
Final volume of the solution = 40 mL + 140 mL
Final volume of the solution = 180 mL
Step 6: Calculate the concentration of the excess HCl.
Concentration of excess HCl = excess moles of HCl / final volume of the solution
Concentration of excess HCl = 0.018 moles / (180 mL / 1000 mL/1 L)
Concentration of excess HCl = 0.1 M
Step 7: Calculate the total concentration of HCl in the final solution.
Total concentration of HCl = concentration of excess HCl + concentration of HCl added
Total concentration of HCl = 0.1 M + 0.2 M
Total concentration of HCl = 0.3 M
Step 8: Calculate the pH.
pH = -log10(total concentration of HCl)
pH = -log10(0.3)
pH ≈ 0.52
Therefore, the pH of the solution after adding 140 mL of HCl is approximately 0.52.