Consider 25.0 ml of .300 MHNO2 that is titrated with .500 M Find ph at eq pt
The pH at the equivalence point is determined by the hydrolysis of the salt. What's the salt? 0.500 M what?
Set up an equation for the hydrolysis, an ICE chart, Kb for NO2^- and solve for OH then convert to pH. Show your work if you get stuck.
okay so i think i did this question correct but i get a little confused with equivalence point. If there is 25 ml of a .300M solution of HNO2...when i set up my ICF table, does that mean there is 7.5 mmol of HNO2 and OH- in the equation HN02 + 0H --> NO2 + H20 since it is at the equivalence pt? Even the the Molarity of NaOH is .500M?
To find the pH at the equivalence point of this titration, we need to determine the concentration of the resulting solution after the reaction has gone to completion. We will calculate the moles of HNO2 and NaOH that react with each other and convert them into the resulting concentration of species.
First, let's determine the number of moles of HNO2 present in the 25.0 ml of 0.300 M HNO2 solution:
moles of HNO2 = volume (in L) × concentration (in M)
= 25.0 ml × (1 L / 1000 ml) × 0.300 M
= 0.0075 moles
Since HNO2 and NaOH react in a 1:1 ratio, the number of moles of NaOH required for complete reaction is also 0.0075 moles.
Now, let's calculate the volume of 0.500 M NaOH solution required to react with 0.0075 moles of NaOH:
volume of NaOH = moles of NaOH / concentration of NaOH
= 0.0075 moles / 0.500 M
= 0.015 L = 15.0 ml
So, at the equivalence point, after the addition of 15.0 ml of 0.500 M NaOH, the volume of the resulting solution will be 25.0 ml + 15.0 ml = 40.0 ml.
Since NaOH is a strong base, it will react completely with HNO2 to form NaNO2 and water. NaNO2 is the salt of a weak acid-strong base reaction and will undergo hydrolysis in water. The resulting solution will be basic.
To find the pH at the equivalence point, we need to consider the dissociation of NaNO2 in water. NaNO2 is the salt of NaOH (strong base) and HNO2 (weak acid).
The hydrolysis of the salt NaNO2 can be represented by the following equation:
NaNO2 + H2O ↔ NaOH + HNO2
Since NaOH is a strong base, it will fully dissociate into Na+ and OH- ions. NaNO2 will partially dissociate into Na+ and NO2- ions, and HNO2 will partially dissociate as a weak acid into H+ and NO2- ions.
At the equivalence point, the concentration of Na+ will be equal to the concentration of OH- ions from the full dissociation of NaOH. The concentration of NO2- ions will be double the concentration of HNO2 from its partial dissociation.
Now, let's calculate the concentrations of Na+, OH-, H+, and NO2- ions. Keep in mind that the volume of the resulting solution at the equivalence point is 40.0 ml.
Concentration of Na+ and OH-: Since we added 15.0 ml of 0.500 M NaOH, the moles of Na+ and OH- are:
moles Na+ = volume (in L) × concentration (in M)
= 15.0 ml × (1 L / 1000 ml) × 0.500 M
= 0.0075 moles
Concentration of Na+ = Concentration of OH- = moles Na+ / volume (in L)
= 0.0075 moles / (40.0 ml × (1 L / 1000 ml))
= 0.1875 M
Concentration of H+ and NO2-: The concentration of HNO2 is 0.300 M, and it partially dissociates. Therefore:
Concentration of H+ = 2 × Concentration of HNO2
= 2 × 0.300 M
= 0.600 M
Concentration of NO2- = Concentration of HNO2
= 0.300 M
At the equivalence point, NaOH and HNO2 have reacted, resulting in the formation of NaNO2, NaOH, HNO2, H+, OH-, NO2- ions in the solution. Since NaOH is a strong base, we can consider the OH- concentration as the hydroxide ion concentration.
To find the pOH at the equivalence point, we use the concentration of OH-:
pOH = -log10(OH- concentration)
pOH = -log10(0.1875) = 0.728
pH + pOH = 14 (at 25°C)
pH = 14 - pOH = 14 - 0.728 ≈ 13.272
Therefore, at the equivalence point of this titration, the pH is approximately 13.272.