Add 0.40 g of NaOH to 50.0 mL of 0.10 M HCOOH, and find approximate pH. No volume change, Ka HCOOH=1.8x10^-4.
I recomputed this and I get pH 5. Right?
Not even close.
HCOOH + NaOH ==> HCOONa + H2O
I think you don't know what you have in the solution. Answer the following. It should lead you to the correct answer.
How many moles HCOOH initially?
How many moles NaOH initially?
How many moles HCOONa are produced?
Which reagent is in excess? my how many moles?
What is the moles/L (the concn) of the excess reagent?
What is the pH?
To find the approximate pH of the solution after adding NaOH to the HCOOH solution, we need to understand the reaction that occurs between these two compounds.
HCOOH (formic acid) is a weak acid, and NaOH is a strong base. When these two compounds react, they undergo a neutralization reaction. The balanced chemical equation for the reaction is:
HCOOH + NaOH -> HCOONa + H2O
This reaction forms HCOONa (sodium formate) and water.
To determine the pH, we need to calculate the concentration of H+ ions present in the resulting solution after the reaction.
Given:
- Mass of NaOH = 0.40 g
- Initial volume of HCOOH = 50.0 mL
- Concentration of HCOOH = 0.10 M
- Ka of HCOOH = 1.8x10^-4
First, let's calculate the moles of NaOH:
Moles of NaOH = mass of NaOH / molar mass of NaOH
The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol.
Moles of NaOH = 0.40 g / 39.99 g/mol = 0.010 mol
Since no volume change is mentioned, the volume of the resulting solution will be 50.0 mL.
Next, let's calculate the moles of HCOOH:
Moles of HCOOH = concentration of HCOOH x volume of HCOOH (in liters)
Since the volume of HCOOH is given in milliliters, we need to convert it to liters:
Volume of HCOOH = 50.0 mL = 50.0 mL x (1 L / 1000 mL) = 0.050 L
Moles of HCOOH = 0.10 M x 0.050 L = 0.005 mol
From the balanced chemical equation, we can see that 1 mol of HCOOH reacts with 1 mol of NaOH to form 1 mol of HCOONa. Therefore, the moles of HCOONa formed will also be 0.005 mol.
Now, let's calculate the concentration of HCOOH and HCOONa in the resulting solution:
Concentration of HCOOH = Moles of HCOOH / Total volume of solution (in liters)
Total volume of solution = Volume of HCOOH + Volume of NaOH
Since there is no volume change mentioned, the total volume of the solution will be 50.0 mL + 50.0 mL = 100.0 mL = 0.100 L
Concentration of HCOOH = 0.005 mol / 0.100 L = 0.050 M
Concentration of HCOONa = 0.005 mol / 0.100 L = 0.050 M
Now, let's calculate the concentration of H+ ions in the solution using the dissociation constant (Ka) of HCOOH:
[H+] = sqrt(Ka x [HCOOH])
[H+] = sqrt((1.8x10^-4 M) x (0.050 M))
[H+] = sqrt(9x10^-6 M^2)
[H+] = 3x10^-3 M
Finally, let's calculate the pH:
pH = -log[H+]
pH = -log(3x10^-3)
pH ≈ 2.52
Therefore, the approximate pH of the resulting solution after adding 0.40 g of NaOH to 50.0 mL of 0.10 M HCOOH is approximately 2.52.
Note: There might be some variations or rounding in the final pH value due to approximations made during the calculation.