You are titrating 35.00 mL of a 1.20 M solution of HCOOH with a strong base. If you add 1.00 mL of a 3.60 M solution of sodium hydroxide, what is the final pH of the acid solution? The Ka for HCOOH is 1.8 X 10-4.
begin with 35.00 mL x 1.20 = 42 mmol.
add 1.00 mL x 3.60 M NaOH = 3.60 mmol.
..........HCOOH + NaOH ==> HCOONa + H2O
begin.....42 mmol...0........0.......0
change....-3.60.....3.60....3.60.....3.60
final......38.40....0.......3.60.....3.60
You see the final solutin is a buffer; i.e., it has a weak acid (HCOOH) and a salt of the weak acid(HCOONa) present.
Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base)/(acid)]
Use pKa of HCOOH
(base) = (HCOONa)
(acid) = (HCOOH).
Plug and chug. Post your work if you get stuck.
If a sample of oxygen-15 has an activity of 8000 becquerels, how many minutes will elapse before it reaches an activity of 500 becquerels?
To find the final pH of the acid solution after adding a strong base, we need to understand the process of neutralization and the calculation of the resulting concentration of the acid and its conjugate base.
First, let's start by writing the balanced chemical equation for the reaction between HCOOH (formic acid) and NaOH (sodium hydroxide):
HCOOH + NaOH -> HCOONa + H2O
According to the stoichiometry of the equation, 1 mole of HCOOH reacts with 1 mole of NaOH to form 1 mole of HCOONa and 1 mole of water.
Next, we need to determine the moles of HCOOH and NaOH that react.
The volume of the HCOOH solution is given as 35.00 mL. To convert this to liters, we divide by 1000:
35.00 mL / 1000 mL = 0.03500 L
The concentration of the HCOOH solution is given as 1.20 M. Using the equation C = n/V, where C is the concentration in moles per liter, n is the number of moles, and V is the volume in liters, we can calculate the number of moles of HCOOH:
n(HCOOH) = C(HCOOH) * V(HCOOH)
= 1.20 M * 0.03500 L
= 0.0420 moles
The volume of the NaOH solution added is given as 1.00 mL. Converting this to liters:
1.00 mL / 1000 mL = 0.00100 L
The concentration of the NaOH solution is given as 3.60 M. Using the same equation as before, we can calculate the number of moles of NaOH:
n(NaOH) = C(NaOH) * V(NaOH)
= 3.60 M * 0.00100 L
= 0.00360 moles
Since the stoichiometry of the balanced equation is 1:1, the number of moles of HCOOH that react is equal to the number of moles of NaOH that react.
After the reaction between HCOOH and NaOH, the HCOOH is partially neutralized, and its concentration decreases. To calculate the new concentration of HCOOH in the final solution, we subtract the number of moles of NaOH reacted from the initial number of moles of HCOOH:
n(HCOOH final) = n(HCOOH initial) - n(NaOH)
= 0.0420 moles - 0.00360 moles
= 0.0384 moles
Now, to find the final concentration of HCOOH, we divide the number of moles of HCOOH by the final volume of the solution:
C(HCOOH final) = n(HCOOH final) / V(final)
= 0.0384 moles / 0.03600 L
= 1.07 M
Next, we need to calculate the concentration of the conjugate base, HCOO⁻ (formate ion). Since the dissociation of HCOOH is in equilibrium, we can use the equation for weak acid dissociation:
Ka = [H⁺][HCOO⁻] / [HCOOH]
The equilibrium concentration of HCOOH is equal to the initial concentration of HCOOH minus the concentration of HCOOH that reacted:
[HCOOH] = C(HCOOH) - C(HCOOH final)
= 1.20 M - 1.07 M
= 0.13 M
Since the stoichiometry of the reaction is 1:1, the concentration of HCOO⁻ is equal to the concentration of the final HCOOH:
[HCOO⁻] = C(HCOOH final)
= 1.07 M
Now, we can rearrange the equation for Ka to solve for [H⁺]:
[H⁺] = (Ka * [HCOOH]) / [HCOO⁻]
= (1.8 x 10^(-4) * 0.13 M) / 1.07 M
= 2.2 x 10^(-5) M
Finally, to find the pH, we need to take the negative logarithm of [H⁺]:
pH = -log10[H⁺]
= -log10(2.2 x 10^(-5) M)
= 4.657
Therefore, the final pH of the acid solution after adding 1.00 mL of a 3.60 M solution of sodium hydroxide is approximately 4.657.