When 1.10 mol of NaOH is added to 1.00 L of 0.30 M HCl the pH of the resulting mixture is:
a) 0.52
b) 0.70
c) 1.00
d) 13.30
Oops 0.10 mol NaOH
0.70
Well, you've just thrown a party with NaOH and HCl! Talk about a chemistry mixer! Now, let's see what happens when they get together.
When 1.10 mol of NaOH and 1.00 L of 0.30 M HCl decide to hang out, some special chemistry magic takes place. The NaOH reacts with HCl to form water (H2O) and NaCl, also known as good old table salt.
Now, the pH of a solution depends on the concentration of H+ ions. In our case, the H+ ions come from the HCl. Since we have an equal amount of NaOH and HCl, they essentially neutralize each other. So, the resulting solution will be close to neutral, which means the pH will be around 7.
But, wait a second! We have to take into account something called the self-ionization of water. Even though our solution is mostly water, it will still have some H+ and OH- ions from the water itself.
The pH scale ranges from 0 to 14. pH values less than 7 are acidic, pH values greater than 7 are basic, and a pH of 7 is neutral.
In this case, the reaction between NaOH and HCl will slightly increase the concentration of OH- ions, making our solution slightly basic. So, the pH would be greater than 7.
Therefore, based on all this chemistry mixing and clown calculations, the closest answer would be d) 13.30. So, the pH of the resulting mixture is close to a classroom full of clowns - definitely a laughter-filled 13.30!
To determine the pH of the resulting mixture, we need to consider the reaction between NaOH (sodium hydroxide) and HCl (hydrochloric acid). The reaction between a strong base and a strong acid produces a neutral salt and water. In this case, NaOH reacts with HCl to form NaCl (sodium chloride) and water (H₂O).
The balanced chemical equation for the reaction is as follows:
NaOH + HCl -> NaCl + H₂O
In this reaction, equal moles of NaOH and HCl react to give equal moles of NaCl and water.
Given:
- Moles of NaOH = 1.10 mol
- Initial concentration of HCl = 0.30 M
- Initial volume of HCl solution = 1.00 L
First, we need to calculate the moles of HCl:
Moles of HCl = Initial concentration of HCl × Initial volume of HCl solution
Moles of HCl = 0.30 M × 1.00 L = 0.30 mol
Since the moles of NaOH and HCl are equal, all of the HCl will react completely with NaOH, leaving no excess HCl. This means that the solution will contain only NaCl and water.
Now, let's calculate the resulting concentration of NaCl:
Moles of NaCl = Moles of HCl = 0.30 mol
Volume of resulting solution = Initial volume of HCl solution = 1.00 L
Concentration of NaCl = Moles of NaCl / Volume of resulting solution
Concentration of NaCl = 0.30 mol / 1.00 L = 0.30 M
The resulting solution will have a concentration of NaCl of 0.30 M. Since NaCl is a neutral salt, it will not affect the pH of the solution.
Therefore, the pH of the resulting mixture will be the same as the pH of the initial HCl solution, which is given as 0.30 M. However, none of the options provided in the question match this pH. Therefore, none of the options is correct.
To find the pH of the resulting mixture when NaOH is added to HCl, we need to determine if the resulting solution is acidic, basic, or neutral.
First, let's calculate the number of moles of HCl present in the solution:
Moles of HCl = Volume (in liters) * Concentration
Moles of HCl = 1.00 L * 0.30 M
Moles of HCl = 0.30 moles
Next, let's calculate the number of moles of NaOH added:
Moles of NaOH = 1.10 moles
Since HCl is a strong acid and NaOH is a strong base, they will react completely in a 1:1 ratio.
Therefore, the moles of HCl that will react with NaOH = 0.30 moles.
After the reaction, we are left with:
Moles of HCl remaining = 0.30 moles - 0.30 moles
Moles of HCl remaining = 0 moles
Since there are no moles of HCl remaining, there is no acidity in the resulting mixture. Therefore, the resulting mixture is basic.
To find the pH of a basic solution, we use the formula:
pOH = -log[OH-]
However, we do not have the concentration of OH- in the resulting mixture. To find it, we can use the fact that NaOH is a strong base and fully dissociates in water, giving one mole of OH- ions per mole of NaOH.
Concentration of OH- = Moles of NaOH / Volume (in liters)
Concentration of OH- = 1.10 moles / 1.00 L
Concentration of OH- = 1.10 M
Now, we can calculate the pOH:
pOH = -log(1.10)
pOH ≈ 0.04
Finally, we can obtain the pH by using the equation:
pH = 14 - pOH
pH = 14 - 0.04
pH ≈ 13.96
Since none of the answer choices match the calculated pH of approximately 13.96, none of the given answer choices are correct. Please recheck the answer choices provided or the calculations done.