What is the pH of a solution when 500ml of 0.2M NaOH and 500ml of 0.2M H2SO4 are mixed? Pls help.
2NaOH + H2SO4 ==> 2H2O + Na2SO4
You have 0.5L x 0.2M = 0.1 mol NaOH
You have 0.5L x 0.2M = 0.1 mol H2SO4.
Which is the limiting reagent? It must be NaOH; i.e., 0.1 mol H2SO4 will take 0.2 mol NaOH for complete neutralization and you don't have that much; therefore, the NaOH will neutralize 1/2 the H2SO4 and leave NaHSO4. The pH will be determined by the ionization of HSO4^-. It was 0.2, half is gone to make it 0.1 and that has been diluted by 1/2 to make the concn = 0.05.
..........HSO4^- ==> H^+ + SO4^2-
initial...0.05M......0.......0
change.....-x.........x.......x
equil....0.05-x.......x.......x
Substitute from the ICE chart into K2 for H2SO4 and solve for x, then convert to pH.
To determine the pH of the solution, we need to consider the reaction between NaOH and H2SO4. When these two strong acids and bases react, they create water (H2O) and a salt (Na2SO4).
First, let's calculate the amount of moles for NaOH and H2SO4:
For NaOH:
Volume = 500 mL = 0.5 L
Molarity = 0.2 M
Moles of NaOH = Molarity x Volume = 0.2 M x 0.5 L = 0.1 moles
For H2SO4:
Volume = 500 mL = 0.5 L
Molarity = 0.2 M
Moles of H2SO4 = Molarity x Volume = 0.2 M x 0.5 L = 0.1 moles
Since the reaction between NaOH (a strong base) and H2SO4 (a strong acid) is a 1:1 reaction, we have equal moles of the two reactants. Therefore, all the NaOH and H2SO4 would react completely, resulting in the formation of 0.1 moles of Na2SO4 and 0.1 moles of H2O.
Now, let's calculate the concentration of the resulting Na2SO4 solution:
Total volume = 500 mL + 500 mL = 1 L
Moles of Na2SO4 = 0.1 moles
Concentration = Moles/Volume = 0.1 moles/1 L = 0.1 M
Since Na2SO4 is a neutral salt, it does not significantly affect the pH of the solution. Therefore, the remaining factor determining the pH is the water.
Water dissociates to form equal amounts of H+ and OH- ions. Since we started with equal amounts of NaOH and H2SO4, the resulting 0.1 moles of H2O would produce 0.1 moles of H+ ions and 0.1 moles of OH- ions.
However, since the solution contains H+ ions (due to the reaction between NaOH and H2SO4), it will be acidic. We need to determine the excess of H+ ions to calculate the pH.
The H+ ions come from the H2O molecules and the remaining H2SO4.
Moles of H+ ions = Initial H2O + Remaining H2SO4
= 0.1 moles (from H2O dissociation) + Moles of H2SO4
Since H2SO4 is a strong acid and completely ionizes, the moles of H+ ions from H2SO4 would be 0.1 moles (same as the initial moles of H2SO4).
So, the total moles of H+ ions in the solution = 0.1 moles + 0.1 moles = 0.2 moles.
To calculate the concentration of H+ ions (pH):
Concentration of H+ ions = Moles of H+ ions/Total volume
= 0.2 moles/1 L
= 0.2 M
The pH of the solution can then be determined using the formula:
pH = -log[H+]
= -log(0.2)
= -(-0.7) (log 0.2 is approximately -0.7)
= 0.7
Therefore, the pH of the solution is approximately 0.7.
To determine the pH of a solution resulting from the mixture of NaOH and H2SO4, you need to consider the reaction that occurs between the two substances. NaOH is a strong base, and H2SO4 is a strong acid.
Step 1: Write the balanced chemical equation for the reaction:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
Based on this equation, you can see that NaOH and H2SO4 react in a 1:1 ratio. This means that all the NaOH will react with an equal amount of H2SO4.
Step 2: Calculate the number of moles of NaOH and H2SO4:
moles of NaOH = (0.2 mol/L) * (0.500 L) = 0.1 moles
moles of H2SO4 = (0.2 mol/L) * (0.500 L) = 0.1 moles
Since the reaction occurs in a 1:1 ratio, 0.1 moles of NaOH react with 0.1 moles of H2SO4.
Step 3: Determine the limiting reactant:
Since the reactants are in equal amounts, neither is in excess. Therefore, neither is a limiting reactant.
Step 4: Calculate the resulting concentration and volume of the salt (Na2SO4):
The resulting concentration of Na2SO4 is 0.1 moles / (0.500 L + 0.500 L) = 0.05 M. The volume is the sum of the volumes of NaOH and H2SO4, which is 500 mL + 500 mL = 1000 mL or 1 L.
Step 5: Calculate the concentration of OH- ions in the solution:
Since NaOH is a strong base, it dissociates completely in water, yielding one hydroxide ion (OH-) for every NaOH molecule. The concentration of OH- ions is therefore 0.05 M.
Step 6: Calculate the concentration of H+ (or H3O+) ions in the solution:
Since H2SO4 is a strong acid, it also dissociates completely in water, yielding two hydrogen ions (H+ or H3O+) for every H2SO4 molecule. The concentration of H+ ions is therefore 0.1 M.
Step 7: Calculate the pH of the solution:
pH = -log[H+]
pH = -log(0.1)
pH ≈ 1
Thus, the pH of the resulting solution is approximately 1.