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This page  contains sample problems and provides the solutions for the problems.

Example: Write the balanced chemical reaction and calculate all of the final concentrations when you add 500 mL of 1.0 M HCl and 500 mL of 1.0 M of NaOH

Solution: This chemical reaction can be seen in the Examples portion of the site. Add in the concentrations and number of moles as well.

HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)

1 M             1 M

0.5 L           0.5 L

0.5 mol        0.5 mol

All of the mole to mole ratios are 1:1, so and [HCl] = [NaOH], so they will neutralize, producing water and a salt. 1.0 L of water will be formed, calculated by adding 500 mL with 500 mL. There will be 0.5 mol of NaCl by the mole to mole ratios, so the concentration of NaCl equals 0.5 mol / 1.0 L which equals 0.5 M.

Solution Stoichiometry

Example: How many moles of water (H2O ) form when 75 milliliters of 0.200M HCl (a strong acid) is completely neutralized with NaOH (a strong base)?

The Steps:

1.      Write out the balanced equation of the reaction.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

2.      Find the mole ratio between the products and the reactants

In this reaction, 1 mole of HCl would yield 1 mole of H2O.

3.      Calculate the number of moles of the starting products

In this reaction, you are given 75 milliliters of 0.200M HCl

(0.075L)(0.200mol/L) = 0.015 mol of HCl

4.      Using mole ratios, calculate the moles of products formed

0.015 mol HCl (1/1) = 0.015 mol H2O

0.015 mol H2O is your answer

Neutralization Reactions

Example: How many milliliters of 0.300M H2SO4 is required to neutralize 50 milliliters of 0.150M KOH?

The Steps:

1.      Look at how the reactants dissociate

Because the reaction involves a strong acid and a strong base, both members dissociate completely. So by observation, 1 mol of H2SO4 would give up 1 mol of H+ atoms, and 1 mol of KOH would give up 1 mol of OH- atoms.

2.      Calculate the mols of the given reactant, in this case: KOH

(0.050L)(0.150 mols/L) = 0.0075 mol KOH, or OH-

3.      Using information found in step 2, calculate the required amount of the other H+ needed.

For each mol of OH- present, 1 mol of H+ is needed to neutralize it. In this case, there are 0.0075 mols of OH- present, which means that 0.0075 mols of H+, or H2SO4 must be needed to neutralize the OH-

4.      Using the molarity given, calculate the amount H2SO4 required to neutralize the KOH.

(x L)(0.300 mol/L) = 0.0075 mol H2SO4

Through division, x = 0.025 L H2SO4, or 25 milliliters H2SO4

25 milliliters H2SO4 is your answer

Related information

Formation of salt:

Salts are usually one of the products in an acid-base neutralization reaction. To make sure that you have the correct chemical formula for the salt, check the charges. Salts are made up of anions and cations, each of which has a charge. For example, with Cl- and Na+, each charge has a magnitude of 1, so the charges cancel it out. However, if you have Mg2+ and Cl-, you need two Cls per Mg in each molecule, so the final chemical formula would be MgCl2.

Now, to get these salts, you look at the acid and base being reacted. In most cases, you can just subtract the H+ ions from the acid and the OH- ions from the bases and put together what’s left over. For example, with HCl and NaOH, remove H+ from HCl and you get Cl-, and remove OH- from NaOH and you get Na+. Then add Cl- and Na+, and you get NaCl.

Formation of a buffer:

A buffer includes a weak acid or base and its conjugate base or acid. A buffer can be formed from a weak acid/base and a salt that includes its conjugate base/acid. For example, acetic acid and sodium acetate will form a buffer.

A buffer will also form when a strong acid or base reacts with an excess of a weak base or acid. The strong agent partially neutralize the weak agent, thus resulting in a weak acid or base and its conjugate.

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