A 50.0 mL aqueous solution of sodium hydroxide has a pH of 12.50. If 36.00 ml of 0.0200 mol/L sulfuric acid is added to this sodium hydroxide solution, what will be the new pH of the resulting solution? Assume that the temperature stays constant at 25 C, and that the volumes are perfectly additive.
I dont understand how what to do? Like do i do IRF or ICE tables or what? Please write a step by step. Thanks
Well, pH can be a bit tricky, but I'll try to explain it step by step for you. And don't worry, I won't clown around with this one!
Step 1: Calculate the initial concentration of hydroxide ions (OH-) in the sodium hydroxide solution.
Since the pH is 12.50, we can use the equation pH = -log[H+]. In this case, [H+] is equal to [OH-] because we are dealing with a strong base. So, to find [OH-], we take 10 to the power of the negative pH value. Therefore, [OH-] = 10^(-12.50).
Step 2: Calculate the number of moles of hydroxide ions (OH-) present in the original 50.0 mL of the sodium hydroxide solution.
To do this, you need to multiply the initial concentration of hydroxide ions ([OH-]) by the volume of the solution in liters (50.0 mL converted to liters).
Step 3: Calculate the number of moles of sulfuric acid (H2SO4) added to the solution.
This can be accomplished by multiplying the volume of sulfuric acid added (36.00 mL converted to liters) by its concentration (0.0200 mol/L).
Step 4: Determine the limiting reagent.
Compare the number of moles of hydroxide ions (OH-) from step 2 with the number of moles of sulfuric acid (H2SO4) from step 3. The reactant that produces fewer moles of product is the limiting reagent.
Step 5: Calculate the number of moles of remaining hydroxide ions (OH-) after the reaction.
Subtract the number of moles of hydroxide ions reacted with the limiting reagent from the total number of moles of hydroxide ions in the original solution.
Step 6: Calculate the volume of the resulting solution.
Add the volumes of the sodium hydroxide solution and the sulfuric acid solution to find the total volume of the resulting solution.
Step 7: Calculate the new concentration of hydroxide ions (OH-) in the resulting solution.
Divide the number of moles of remaining hydroxide ions by the volume of the resulting solution to obtain the new concentration.
Step 8: Calculate the new pH of the resulting solution.
To find the new pH, use the equation pH = -log[H+]. Once again, we can assume [H+] is equal to [OH-] for a strong base.
I hope this step-by-step guide helps you solve the problem. Remember, don't be afraid to ask for help if you need it. And always remember, laughing at chemistry problems is always a good solution!
To find the new pH of the resulting solution, you can use the concept of neutralization. Here's a step-by-step process to solve the problem:
Step 1: Calculate the concentration of sodium hydroxide (NaOH) in the original solution.
- The pH of the solution is 12.50, which means that the concentration of hydroxide ions (OH-) is 10^(-pH).
- Calculate the concentration of OH-: [OH-] = 10^(-12.50) M.
Step 2: Determine the number of moles of NaOH in the original solution.
- The volume of the solution is given as 50.0 mL or 0.050 L.
- Use the equation: moles = concentration × volume.
- Moles of NaOH = [OH-] × volume of solution = (10^(-12.50)) × 0.050.
Step 3: Calculate the concentration of sulfuric acid (H2SO4) in the added solution.
- The volume of H2SO4 added is 36.00 mL or 0.036 L.
- The concentration of H2SO4 is given as 0.0200 mol/L.
- Calculate the number of moles of H2SO4: moles = concentration × volume.
Step 4: Determine the limiting reagent.
- Compare the number of moles of NaOH (from Step 2) and H2SO4 (from Step 3) to identify the limiting reagent.
- The limiting reagent is the one with fewer moles.
Step 5: Calculate the number of moles of the excess reagent left over.
- Subtract the moles of the limiting reagent from the initial moles of the excess reagent.
- The excess reagent is the one that is not limiting.
- Determine the moles of excess NaOH or H2SO4 remaining.
Step 6: Determine the new volume of the resulting solution.
- The volume of the original solution is 50.0 mL or 0.050 L.
- Add the volume of the added solution, which is 36.00 mL or 0.036 L.
- Calculate the total volume of the solution.
Step 7: Calculate the new concentration of the OH- ion.
- Determine the moles of OH- remaining.
- Divide the moles by the total volume of the resulting solution to find the new concentration of OH-.
Step 8: Calculate the new pOH.
- Use the new concentration of OH- from the previous step.
- Calculate pOH: pOH = -log[OH-] (logarithm with base 10).
Step 9: Calculate the new pH.
- Use the pOH value from the previous step to calculate the new pH.
- pH + pOH = 14.00 (for water at 25 °C).
- Calculate pH: pH = 14.00 - pOH.
Follow these steps to solve the problem, plugging in the values given in the question.
To determine the new pH of the resulting solution, you need to follow these steps:
Step 1: Calculate the moles of sodium hydroxide (NaOH) in the initial solution.
The molarity of the sodium hydroxide solution is not given, so we need to use the pH to determine the concentration of hydroxide ions (OH-) in the solution. The pH of a solution is given by the equation: pH = -log[H+].
Since we are given the pH, we can calculate the concentration of hydrogen ions ([H+]):
[H+] = 10^(-pH)
[H+] = 10^(-12.50) = 3.16 x 10^(-13) mol/L
Since NaOH is a strong base and it completely dissociates in water, the concentration of hydroxide ions ([OH-]) will be the same as the concentration of sodium hydroxide (NaOH):
[OH-] = [NaOH] = 3.16 x 10^(-13) mol/L
To calculate the moles of NaOH in the solution, we use the formula:
moles = concentration (mol/L) * volume (L)
moles of NaOH = 3.16 x 10^(-13) mol/L * 0.0500 L = 1.58 x 10^(-14) mol
Step 2: Determine the moles of sulfuric acid (H2SO4) added.
The volume of sulfuric acid added is given as 36.00 mL, and the molarity is 0.0200 mol/L. Using the moles formula:
moles of H2SO4 = concentration (mol/L) * volume (L)
moles of H2SO4 = 0.0200 mol/L * 0.0360 L = 7.20 x 10^(-4) mol
Step 3: Determine the excess reactant and calculate the moles of hydroxide ions remaining.
Since sulfuric acid is a strong acid and completely dissociates in water, it will react with all the hydroxide ions in the NaOH solution. To determine the excess reactant, we compare the moles of NaOH and H2SO4:
Moles of NaOH = 1.58 x 10^(-14) mol
Moles of H2SO4 = 7.20 x 10^(-4) mol
Since Moles of NaOH < Moles of H2SO4, NaOH is the limiting reagent. This means that all the NaOH will react, and some H2SO4 will be left.
To calculate the remaining moles of H2SO4, we subtract the moles of NaOH reacted from the initial moles of H2SO4:
Remaining moles of H2SO4 = Moles of H2SO4 - Moles of NaOH
Remaining moles of H2SO4 = 7.20 x 10^(-4) mol - 1.58 x 10^(-14) mol = 7.20 x 10^(-4) mol (approximately)
Therefore, the moles of hydroxide ions remaining = moles of NaOH - moles of H2SO4
moles of hydroxide ions = 1.58 x 10^(-14) mol - 7.20 x 10^(-4) mol ≈ -7.19 x 10^(-4) mol
Note: The negative sign indicates that there is an excess of H2SO4, and the concentration of hydroxide ions will be negative.
Step 4: Calculate the new concentration of hydroxide ions and the new pOH.
The concentration of hydroxide ions ([OH-]) is given by:
[OH-] = moles of hydroxide ions / volume (L)
Since we have the volume of the initial solution (50.0 mL), and we added the volume of sulfuric acid (36.00 mL) to it, the total volume of the resulting solution is:
Total volume = 50.0 mL + 36.00 mL = 86.00 mL = 0.0860 L
Therefore, the new concentration of hydroxide ions is:
[OH-] = moles of hydroxide ions / volume (L)
[OH-] = -7.19 x 10^(-4) mol / 0.0860 L ≈ -8.37 x 10^(-3) mol/L
The pOH is calculated using the equation: pOH = -log[OH-]:
pOH = -log(-8.37 x 10^(-3)) ≈ 2.08
Step 5: Calculate the new pH.
The pH can be calculated using the equation: pH + pOH = 14
pH = 14 - pOH
pH = 14 - 2.08 ≈ 11.92
Therefore, the new pH of the resulting solution is approximately 11.92.
Note: The negative concentration of hydroxide ions and the resulting pH below 7 indicate that the resulting solution will be acidic due to the excess sulfuric acid added.
What's an IRF?
2NaOH + H2SO4 = Na2SO4 + 2H2O
Convert pH of NaOH to pOH and calculate (OH^-) and (NaOH). Then mols NaOH = M x L = ?
mols H2SO4 = M x L = ?
Subtract to see which is in excess, calculate (H^+) or (OH^-) as it applies and convert to pH.