If 25 mL of 0.20M NaOH is added to 20 mL of 0.25M boric acid, what is the pH of the resultant solution?
Well, looks like NaOH and boric acid are having a little party in the lab! Let's join in on the fun and calculate the pH of their resultant solution.
To start, we need to determine how many moles of NaOH and boric acid we have. Moles, not mold, although I'm sure NaOH and boric acid are quite yucky on bread.
So, for NaOH, we multiply the volume (25 mL) by the molarity (0.20 M) to find the number of moles. That gives us 5 x 10^-3 moles.
For boric acid, we do the same thing. Take the volume (20 mL) and multiply it by the molarity (0.25 M). That gives us 5 x 10^-3 moles as well.
Now, the fun part begins! We mix these two substances together and they undergo a delightful reaction. NaOH (a strong base) reacts with boric acid (a weak acid) to form sodium borate, a.k.a. a new compound.
Since NaOH is a strong base, it completely dissociates in solution, releasing OH- ions. Boric acid, being a weak acid, only partially dissociates, giving us some H+ ions.
The OH- ions and H+ ions then engage in a furious battle, forming water. H2O, the ultimate peacekeeper.
However, some of the OH- ions and H+ ions will still be left over, and they determine the pH. They're like the last two fighters standing in the ring.
Since the moles of NaOH and boric acid are equal, we know that the leftover OH- and H+ ions will also be equal. This is because they react in a 1:1 ratio. How fair and balanced!
Knowing that OH- reacts with H+ in a 1:1 ratio, we can calculate the concentration of leftover H+ ions. They'll be equal to the concentration of OH- ions left over, which is 5 x 10^-3 M.
Now, we can use the equation pH = -log[H+], where [H+] represents the concentration of leftover H+ ions, to find the pH.
Taking the negative logarithm of 5 x 10^-3 M gives us a pH value of 2.3.
So, the pH of the resultant solution is 2.3. Now I hope NaOH and boric acid enjoy their peaceful water-filled solution!
To determine the pH of the resultant solution, we can use the concept of acid-base neutralization and the dissociation of boric acid.
Step 1: Write the balanced equation representing the neutralization reaction between NaOH and H3BO3 (boric acid):
NaOH + H3BO3 → NaH2BO3 + H2O
Step 2: Determine the limiting reagent:
To determine the limiting reagent, we need to compare the number of moles of NaOH and H3BO3.
The number of moles (n) can be calculated using the formula:
n = C × V
Where C is the concentration and V is the volume.
For NaOH:
n(NaOH) = 0.20 mol/L × 0.025 L
n(NaOH) = 0.005 mol
For H3BO3:
n(H3BO3) = 0.25 mol/L × 0.020 L
n(H3BO3) = 0.005 mol
Since both the reactants have the same number of moles, neither NaOH nor H3BO3 will be the limiting reagent. Therefore, the reaction will go to completion, and we can assume complete neutralization.
Step 3: Determine the concentration of the resulting solution:
Since both reactants completely react, the volume of the resulting solution will be the sum of the volumes of NaOH and boric acid:
V(resulting solution) = V(NaOH) + V(H3BO3)
V(resulting solution) = 0.025 L + 0.020 L
V(resulting solution) = 0.045 L
To determine the concentration (C) of the resulting solution, we can use the formula:
C = n/V
For NaH2BO3:
n(NaH2BO3) = n(NaOH) = 0.005 mol
C(NaH2BO3) = 0.005 mol / 0.045 L
C(NaH2BO3) = 0.111 mol/L (or M)
Step 4: Calculate the pH of the resulting solution:
Since NaH2BO3 is a salt, it will hydrolyze in water and produce basic OH- ions. To determine the pH, we need to calculate the pOH first. The pOH can be calculated using the formula:
pOH = -log10(OH- concentration)
For NaH2BO3:
OH- concentration = 2 × C(NaH2BO3) = 2 × 0.111 mol/L
OH- concentration = 0.222 mol/L
pOH = -log10(0.222)
pOH ≈ 0.652
Finally, we can calculate the pH using the equation:
pH + pOH = 14
pH + 0.652 = 14
pH ≈ 14 - 0.652
pH ≈ 13.348
Therefore, the pH of the resultant solution is approximately 13.348.
To determine the pH of the resultant solution after adding NaOH to boric acid, we need to take into account the reaction between the two compounds and the resulting concentration of the acidic and basic species.
Step 1: Determine the moles of boric acid and NaOH.
Moles of boric acid = volume (L) × molarity (mol/L) = 0.020 L × 0.25 mol/L = 0.005 mol
Moles of NaOH = volume (L) × molarity (mol/L) = 0.025 L × 0.20 mol/L = 0.005 mol
Step 2: Write the balanced equation for the reaction between boric acid (H3BO3) and NaOH:
H3BO3 + NaOH → NaH2BO3 + H2O
Step 3: Determine the limiting reactant.
Since both boric acid and NaOH are in a 1:1 ratio, they are also the limiting reactants. This means that all the boric acid and NaOH will react completely.
Step 4: Determine the concentration of the acidic and basic species in the resultant solution.
The moles of the acidic species (boric acid) remaining after the reaction will be zero, as it has reacted completely. The moles of the basic species (NaOH) remaining after the reaction will also be zero. Instead, the resulting solution will contain the product NaH2BO3.
Step 5: Calculate the concentration of NaH2BO3.
To calculate the concentration (molarity) of NaH2BO3 in the resultant solution, we need to know the final volume. Since no volume is specified in the question, we assume that the volumes of the two solutions add up to give a total volume of 45 mL (25 mL + 20 mL).
Final volume (L) = 45 mL × (1 L/1000 mL) = 0.045 L
Molarity (M) = moles of NaH2BO3 / final volume (L)
= 0.005 mol / 0.045 L
= 0.111 M
Step 6: Calculate the pOH of the NaH2BO3 solution.
pOH = -log10[OH-]
As NaH2BO3 is a weak base, we can assume it dissociates into Na+ and H2BO3-.
Since NaOH is a strong base, it will dissociate completely into Na+ and OH-.
Thus, the concentration of OH- ions in the resultant solution will be equal to the concentration of unreacted NaOH, which is 0 M.
pOH = -log10[0] = undefined
Step 7: Calculate the pH of the resultant solution.
Since pH + pOH = 14, and the pOH is undefined in this case, the pH of the resultant solution will also be undefined.
Therefore, the resultant solution after adding NaOH to boric acid does not have a pH value.