Overview Of Acids And Bases - Chemistry LibreTexts

Neutralization

A special property of acids and bases is their ability to neutralize the other's properties. In an acid-base (or neutralization) reaction, the H+ ions from the acid and the OH- ions from the base react to create water (H2O). Another product of a neutralization reaction is an ionic compound called a salt. Therefore, the general form of an acid-base reaction is:

acid + base \(\ce{->}\) water + salt

The following are examples of neutralization reactions:

1. \(\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)}\)

2. \(\ce{2HBr(aq) + Ca(OH)2(aq) -> 2H2O(l) + CaBr2(aq)}\)

Titrations

Titrations are performed with acids and bases to determine their concentrations. At the equivalence point, the number of moles of the acid will equal the number of moles of the base. This indicates that the reaction has been neutralized.

\[\text{moles of acid} = \text{moles of base} \nonumber\]

Here's how the calculations are done:

For instance, hydrochloric acid is titrated with sodium hydroxide:

\(\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)}\)

\(\ce{H+ + OH- -> H2O}\)

\[\text{moles of HCl} = \text{moles of NaOH} \nonumber\]

Example \(\PageIndex{1}\)

30 mL of 1.00 M NaOH is needed to titrate 60 mL of an \(\ce{HCl}\) solution. The concentration of \(\ce{HCl}\) needs to be determined.

Solution

At the equivalence point:

\[\text{moles of HCl} = \text{moles of NaOH} \nonumber\]

\[\begin{gathered} \left(\text { Molarity }_{\text {acid }}\right)\left(\text { Volume }_{\text {acid }}\right)=\left(\text { Molarity }_{\text {base }}\right)\left(\text { Volume }_{\text {base }}\right) \\ \left(\text { Molarity }_{\mathrm{HCl}}\right)\left(\text { Volume }_{\mathrm{HCl}}\right)=\left(\text { Molarity }_{\mathrm{NaOH}}\right)\left(\text { Volume }_{\mathrm{NaOH}}\right) \end{gathered}\]

To solve for the molarity of HCl, plug in the given data into the equation above.

MHCl(60 mL HCl) = (1.00 M NaOH)(30 mL NaOH)

MHCl=0.5 M

The concentration of HCl is 0.5 M.

Exercise \(\PageIndex{1}\)

Which of the following compounds is a strong acid?

1. CaSO4
2. NaCl
3. HNO3
4. NH3

Answer

There are six strong acids and all other acids are considered weak. HNO3 is one of those 6 strong acids, while NH3 is actually a weak base.

The answer is (3) HNO3.

Exercise \(\PageIndex{2}\)

Which of the following compounds is a Brønsted-Lowry base?

1. HCl
2. HPO42-
3. H3PO4
4. NH4+
5. CH3NH3+

Answer

A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H+. This eliminates HCl, H3PO4 ,NH4+ and CH3NH3+ because they are Brønsted-Lowry acids. They all give away protons. In the case of HPO42-, consider the following equation:

\(\ce{HPO4^{2-} + H2O -> PO4^{3-} + H3O+}\)

Here, it is clear that HPO42- is the acid since it donates a proton to water to make H3O+ and PO43-. Now consider the following equation:

\(\ce{HPO4^{2-} + H2O -> H2PO4^{-} + OH-}\)

In this case, HPO42- is the base since it accepts a proton from water to form H2PO4- and OH-. Thus, HPO42- is an acid and base together, making it amphoteric.

Since HPO42- is the only compound from the options that can act as a base, the answer is (2) HPO42-.

Exercise \(\PageIndex{3}\)

A 50 ml solution of 0.5 M NaOH is titrated until neutralized into a 25 ml sample of HCl. What was the concentration of the HCl?

Answer

Since the number of moles of acid equals the number of moles of base at neutralization, the following equation is used to solve for the molarity of HCl:

Now, plug into the equation all the information that is given:

MHCl(25 mL HCl) = (0.5 MNaOH)(50 mL NaOH)

MHCl = 1

The correct answer is 1 MHCl.

Exercise \(\PageIndex{4}\)

In the following acid-base neutralization, 2.79 g of the acid HBr (80.91g/mol) neutralized 22.72 mL of a basic aqueous solution by the reaction:

\[\ce{HBr(aq) + NaOH(aq) \rightarrow H2O(l) + NaBr(s)} \nonumber \]

Answer

First, the number of moles of the acid needs to be calculated. This is done by using the molar mass of HBr to convert 2.79 g of HBr to moles.

(2.79 g HBr)/(80.91 g/mol HBr) = 0.0345 moles HBr

Since this is a neutralization reaction, the number of moles of the acid (HBr) equals the number of moles of the base (NaOH) at neutralization:

moles of acid = moles of base

0.0345 moles HBr = 0.0345 moles NaOH

The molarity of NaOH can now be determined since the amount of moles are found and the volume is given. Convert 22.72 mL to Liters first since molarity is in units of moles/L.

Molarity = (0.0345 moles NaOH)/(0.02272 L NaOH) = 1.52 MNaOH

The correct answer is 1.52 MNaOH.

Exercise \(\PageIndex{5}\)

Which of the following is a Brønsted-Lowry base but not an Arrhenius base?

1. NH3
2. NaOH
3. Ca(OH)2
4. KOH

Answer

The Brønsted-Lowry definition says that a base accepts protons (H+ ions). NaOH, Ca(OH)2, and KOH are all Arrhenius bases because they yield the hydroxide ion (OH-) when they ionize. However, NH3 does not dissociate in water like the others. Instead, it takes a proton from water and becomes NH4 while water becomes a hydroxide.

\(\ce{NH3(aq) + H2O(l) -> NH4(aq)+ + OH-(aq)}\)

Therefore, the correct answer is (1) NH3.

References

1. Brent, Lynnette. Acids and Bases. New York, NY: Crabtree Pub., 2009. Print.
2. Hulanicki, Adam. Reactions of Acids and Bases in Analytical Chemistry. Ellis Horwood Limited: 1987.
3. Oxlade, Chris. Acids & Bases. Chicago, IL: Heinemann Library, 2002. Print.
4. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications. Macmillian: 2007.
5. Vanderwerd, Calvin A. Acids, Bases, and the Chemistry of the Covalent Bond. Reinhold: 1961.