A buffer was made by mixing 0.1522 moles of HCO2H with 0.1832 moles LiHCO2- and diluting to exactly 1L. What will be the pH after addition of 10ml of 0.3657M NaOH to 50mL of the buffer. Ka(HCO2H) = 1.8*10-4
In 50 mL of the buffer you will have
millimoles COO- = 50 mL x 0.18322M = 9.160
mmoles HCOOH = 50mL x 0.1522 = 7.610
mmols NaOH added = 10 mL x 0.3657 = 3.657
...........HCOOH + OH^= ==> HCOO^- + H2O
initial....7.610....0........9.160
add................3.657............
change.....-3.657..-3.657..+3.657..3.657
equil..... sum.......0......sum......sum
Substitute into the HH equation and solve for pH.
To find the pH after the addition of NaOH to the buffer, we need to consider the acid-base reactions that occur.
Step 1: Identify the components of the buffer solution.
The buffer solution is made up of two components: HCO2H (formic acid) and LiHCO2- (sodium formate).
Step 2: Determine the initial concentrations of the buffer components.
Given that the moles of HCO2H is 0.1522 and moles of LiHCO2- is 0.1832, we can convert these values to molarity by dividing by the total volume of the buffer solution (1 L). Therefore, the initial concentrations of HCO2H and LiHCO2- are:
[HCO2H] = 0.1522 moles / 1 L = 0.1522 M
[LiHCO2-] = 0.1832 moles / 1 L = 0.1832 M
Step 3: Calculate the concentrations of the buffer components after dilution.
Since the buffer solution is diluted to 1 L by adding solvent, the concentrations of HCO2H and LiHCO2- will remain the same.
Step 4: Determine the moles of NaOH added.
The volume of NaOH added is given as 10 mL, which can be converted to liters by dividing by 1000. Therefore, the volume of NaOH added is 10 mL / 1000 mL/L = 0.01 L.
The molarity of NaOH added is given as 0.3657 M. Using the volume and molarity, we can calculate the moles of NaOH added:
NaOH moles = 0.01 L * 0.3657 M = 0.003657 moles
Step 5: Determine the concentration change of the components after the reaction.
NaOH reacts with HCO2H, forming HCO2- and water. The stoichiometry of this reaction is 1:1 (1 mole of NaOH reacts with 1 mole of HCO2H). Therefore, the concentration of HCO2H will decrease by the same amount as the moles of NaOH added.
[HCO2H] change = -0.003657 moles / 1 L = -0.003657 M
Since the initial concentration of HCO2H is 0.1522 M, the final concentration of HCO2H after the reaction will be:
[HCO2H] final = 0.1522 M + (-0.003657 M) = 0.148542 M
Step 6: Calculate the concentration change of HCO2-.
Since NaOH reacts with HCO2H in a 1:1 ratio, the concentration of HCO2- will increase by the same amount as the moles of NaOH added.
[HCO2-] change = 0.003657 moles / 1 L = 0.003657 M
Since the initial concentration of LiHCO2- is 0.1832 M, the final concentration of LiHCO2- after the reaction will be:
[LiHCO2-] final = 0.1832 M + 0.003657 M = 0.186857 M
Step 7: Calculate the concentration of OH- ions after the reaction.
The concentration of OH- ions can be calculated from the moles of NaOH added and the total volume of the solution after the reaction. The total volume is the sum of the initial volume of the buffer solution (50 mL) and the volume of NaOH added (10 mL), both converted to liters:
Total volume = (50 mL + 10 mL) / 1000 mL/L = 0.06 L
[OH-] = NaOH moles / Total volume = 0.003657 moles / 0.06 L = 0.06095 M
Step 8: Calculate the concentration ratio of HCO2H to HCO2- after the reaction.
The Henderson-Hasselbalch equation can be used to calculate the pH of the buffer solution after the reaction:
pH = pKa + log([HCO2-] / [HCO2H])
Given that Ka(HCO2H) = 1.8 * 10^-4, we can determine pKa:
pKa = -log(Ka) = -log(1.8 * 10^-4) = 4.744
Substituting the values obtained:
pH = 4.744 + log(0.186857 M / 0.148542 M)
Step 9: Calculate the pH.
Calculate the concentration ratio [HCO2-] / [HCO2H]:
[HCO2-] / [HCO2H] = 0.186857 M / 0.148542 M = 1.256
Substitute the values into the Henderson-Hasselbalch equation:
pH = 4.744 + log(1.256)
Using a calculator, the final pH after addition of NaOH to the buffer is approximately 4.899.