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.