A 100.0-mL sample of 1.00 M #HCl# is titrated with 1.00 M #NaOH#. What is the pH of the solution after 100.0 mL of #NaOH# have been added to the acid?

Answer 1

#"pH" = 7#

This titration involves hydrochloric acid, #"HCl"#, a strong acid, and sodium hydroxide, #"NaOH"#, a strong base, so right from the start you should expect to have a neutral solution at equivalence point.

In other words, when the solution contains enough strong base to neutralize all the acid, its pH should be equal to that of pure water.

Hydrochloric acid will ionize completely in aqueous solution to produce hydronium cations, #"H"_3"O"^(+)#

#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#

Similarly, sodium hydroxide will ionize completely in aqueous solution to produce hydroxide anions, #"OH"^(-)#

#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

When these two solutions are mixed, the hydronium cations and the hydroxide anions will neutralize each other to produce water

#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#

As you can see, it takes equal numbers of moles of hydronium cations and hydroxide anions to have a complete neutralization.

In your case, you have #"100.0 mL"# of #"1.0 M"# hydrochloric acid solution. You're titrating this with #"100.0 mL"# of #"1.0 M"# sodium hydroxide solution.

Since both solution have the same volume and the same concentration, it follows that they also have the same number of moles of hydronium cations and hydroxide anions.

Therefore, mixing these two solutions will result in a complete neutralization.

As a result, the pH of the solution will be equal to that of water, which at room temperature is #7#.

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Answer 2

To find the pH of the solution after 100.0 mL of NaOH have been added to the acid, we need to determine the amount of excess base (NaOH) present in the solution and then calculate the pH based on the resulting concentration of HCl.

Given that the initial volume of HCl is 100.0 mL and its concentration is 1.00 M, the initial moles of HCl can be calculated as follows:

Initial moles of HCl = volume (in liters) * concentration = (100.0 mL / 1000 mL/L) * 1.00 M = 0.100 moles

Since the reaction between HCl and NaOH is a 1:1 stoichiometric ratio, the moles of NaOH added will be equal to the moles of HCl neutralized. Therefore, after 100.0 mL of NaOH (1.00 M) have been added, the moles of NaOH added will be:

Moles of NaOH added = volume (in liters) * concentration = (100.0 mL / 1000 mL/L) * 1.00 M = 0.100 moles

Since the same amount of moles of NaOH have been added as the initial moles of HCl, all of the HCl has been neutralized, leaving only the excess NaOH.

The excess moles of NaOH can be calculated as:

Excess moles of NaOH = moles of NaOH added - moles of HCl initially = 0.100 moles - 0.100 moles = 0 moles

This means that after the titration, there is excess NaOH in the solution, and the pH will be determined by the concentration of the excess NaOH. To find the concentration of excess NaOH, we can use the remaining volume of the solution and the initial volume and concentration of NaOH.

The remaining volume of the solution after titration is 100.0 mL + 100.0 mL = 200.0 mL.

The concentration of excess NaOH can be calculated as:

Concentration of excess NaOH = moles of excess NaOH / volume of solution (in liters) = 0 moles / (200.0 mL / 1000 mL/L) = 0 M

Since the concentration of excess NaOH is 0 M, the solution is basic. Therefore, the pH of the solution is greater than 7, but the exact pH cannot be calculated without knowing the concentration of the excess NaOH.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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