Two aqueous solutions are mixed: 50.0 mL of 1.60×10−2 M H2SO4 and 50.0 mL of 3.75×10−2 M NaOH.

what is the PH of the following solution?

so what the h is the answer

To determine the pH of the resulting solution, we need to calculate the concentration of H+ ions.

Step 1: Calculate the moles of H2SO4:
Molarity (M) = moles/liter
Volume (L) = 50.0 mL = 0.0500 L
Moles of H2SO4 = Molarity * Volume
Moles of H2SO4 = (1.60 × 10^-2 M) * (0.0500 L)

Step 2: Calculate the moles of NaOH:
Molarity (M) = moles/liter
Volume (L) = 50.0 mL = 0.0500 L
Moles of NaOH = Molarity * Volume
Moles of NaOH = (3.75 × 10^-2 M) * (0.0500 L)

Step 3: Determine the limiting reagent:
The limiting reagent is the one with fewer moles. In this case, it is H2SO4 because it produces only one H+ ion per molecule.

Step 4: Calculate the moles of H+ ions:
Moles of H+ ions = Moles of H2SO4

Step 5: Determine the new volume of the resulting solution:
The two solutions are mixed, so the combined volume is:
Total volume = 50.0 mL + 50.0 mL = 100.0 mL = 0.100 L

Step 6: Calculate the new molarity of the H+ ions:
Molarity = Moles of H+/Volume
Molarity = Moles of H+/0.100 L

Step 7: Calculate the pH of the solution:
pH = -log10(Molarity)

By following these steps, you can calculate the pH of the solution.

To determine the pH of the solution formed by mixing H2SO4 and NaOH, we need to consider the reaction that takes place between these two compounds. In this case, sulfuric acid (H2SO4) reacts with sodium hydroxide (NaOH) to form water (H2O) and sodium sulfate (Na2SO4):

H2SO4 + 2NaOH → 2H2O + Na2SO4

Since H2SO4 and NaOH are strong acids and bases, respectively, their reaction is complete. The resulting solution contains water and the salt, sodium sulfate (Na2SO4), which will not affect the pH.

First, we need to determine the concentration of the individual ions in the mixed solution by using the equation:

moles of solute = concentration × volume

For 50.0 mL of 1.60×10−2 M H2SO4:
moles of H2SO4 = (1.60×10−2 mol/L) × (50.0 mL / 1000 mL/mL) = 8.00×10−4 mol

For 50.0 mL of 3.75×10−2 M NaOH:
moles of NaOH = (3.75×10−2 mol/L) × (50.0 mL / 1000 mL/mL) = 1.88×10−3 mol

Since the reaction stoichiometry between H2SO4 and NaOH is 1:2, the amount of H2SO4 consumed will be double that of NaOH. Thus, the remaining moles of H2SO4 are:

moles of H2SO4 remaining = 8.00×10−4 mol - 2(1.88×10−3 mol) = 8.00×10−4 mol - 3.76×10−3 mol = -2.96×10−3 mol

Since the concentration and volume of the final solution remain the same (100.0 mL), we can calculate the new concentration of H2SO4:

new concentration = (moles of solute) / (volume of solution)
new concentration = (-2.96×10−3 mol) / (100.0 mL / 1000 mL/mL) = -2.96×10−2 M

The negative concentration indicates that there is an excess of NaOH, meaning the solution will be basic with a high pH.

Therefore, the pH of this solution is greater than 7, indicating it is basic.

The trick to these question is to recognize what you have in the final solution.

Write the equation.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O

moles H2SO4 = M x L = ??
moles NaOH = M x L= ??
moles H2SO4 required to use all of it will be moles H2SO4 x 2 (since there are two moles NaOH for every mole of H2SO4). Look at those numbers, see which is in excess and use that to calculate either the OH or the H depending upon the one in excess.

Post your work if you get stuck.