What will be the volume of 0.1 m H2SO4 solution required to neutralize 40 mL of 0.1 M KOH solution?

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When an acid and a base react with each other, a neutralization reaction occurs, forming a salt and water. The water forms from the combination of the H+ ions from the acid and the OH- ions from the base. Strong acids and strong bases completely dissociate, so the reaction yields a solution with a neutral pH (pH = 7). Because of the complete dissociation of strong acids and bases, if you're given a concentration of an acid or base, you can determine the volume or quantity of the other chemical required to neutralize it. This example problem explains how to determine how much acid is needed to neutralize a known volume and concentration of a base.

Solving a chemistry problem where a strong acid neutralizes a strong base is straightforward because both the acid and the base completely dissociate.
In contrast, neutralization involving a weak acid and/or a weak base requires that you know and use the dissociation constant.
Neutralization occurs at the point where the number of moles of H+ equals the number of moles of OH-.

Neutralization relies on dissociation of an acid and a base. Dissociation is where the acid or base breaks into its component ions. The ions participating in a neutralization reaction are the H+ from the acid and the OH- from the base. The general form of the reaction is:

acid + base → salt + waterAH + B → A + BH

As an example, when hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), it produces table salt or sodium chloride (NaCl) and water:

HCl + NaOH → NaCl + H2O

Neutralization requires equal amounts of H+ and OH-. So, knowing the volume and concentration of either the acid or base lets you find the volume and concentration of its partner in the reaction.

What volume of 0.075 M HCl is required to neutralize 100 milliliters of 0.01 M Ca(OH)2 solution?

HCl is a strong acid and will dissociate completely in water to H+ and Cl-. For every mole of HCl, there will be one mole of H+. Since the concentration of HCl is 0.075 M, the concentration of H+ will be 0.075 M.

Ca(OH)2 is a strong base and will dissociate completely in water to Ca2+ and OH-. For every mole of Ca(OH)2 there will be two moles of OH-. The concentration of Ca(OH)2 is 0.01 M so [OH-] will be 0.02 M.

So, the solution will be neutralized when the number of moles of H+ equals the number of moles of OH-.

Step 1: Calculate the number of moles of OH-.
Molarity = moles/volume
moles = Molarity x Volume
moles OH- = 0.02 M/100 milliliters
moles OH- = 0.02 M/0.1 liters
moles OH- = 0.002 moles
Step 2: Calculate the Volume of HCl needed
Molarity = moles/volume
Volume = moles/Molarity
Volume = moles H+/0.075 Molarity
moles H+ = moles OH-
Volume = 0.002 moles/0.075 Molarity
Volume = 0.0267 Liters
Volume = 26.7 milliliters of HCl

26.7 milliliters of 0.075 M HCl is needed to neutralize 100 milliliters of 0.01 Molarity Ca(OH)2 solution.

The most common mistake people make when performing this calculation is not accounting for the number of moles of ions produced when the acid or base dissociates. It's easy to understand: only one mole of hydrogen ions is produced when hydrochloric acid dissociates, yet also easy to forget it's not a 1:1 ratio with the number of moles of hydroxide released by calcium hydroxide (or other bases with divalent or trivalent cations).

The other common mistake is a simple math error. Make sure you convert milliliters of solution to liters when you calculate the molarity of your solution!

Skoog, D.A; West, D.M.; Holler, J.F.; Crouch, S.R. (2004). Fundamentals of Analytical Chemistry (8th ed.). Thomson Brooks/Cole. ISBN 0-03-035523-0.
Snoeyink, V.L.; Jenkins, D. (1980). Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters. New York: Wiley. ISBN 0-471-51185-4.
Trummal, Aleksander; Lipping, Lauri; Kaljurand, Ivari; Koppel, Ilmar A.; Leito, Ivo (2016). "Acidity of Strong Acids in Water and Dimethyl Sulfoxide". The Journal of Physical Chemistry A. 120 (20): 3663–3669. doi:10.1021/acs.jpca.6b02253
Zumdahl, Steven S. (2009). Chemical Principles (6th ed.). New York: Houghton Mifflin Company.

Chemistry Solutions Molarity

Now the #"moles of solute"# are a constant. The volume of solution MAY change substantially with increasing or decreasing temperature. In some calculations #"molality"# is used in preference, which is defined by the quotient....

#"molality"="moles of solute"/"kilograms of solvent"#

....this is temperature independent, and at lower concentrations, #"molarity"-="molality"#.
To get the molarity, you divide the moles of solute by the litres of solution.

#"Molarity" = "moles of solute"/"litres of solution"#

For example, a 0.25 mol/L NaOH solution contains 0.25 mol of sodium hydroxide in every litre of solution.

To calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution.

To calculate molarity:
1. Calculate the number of moles of solute present.
2. Calculate the number of litres of solution present.
3. Divide the number of moles of solute by the number of litres of solution.
EXAMPLE:

What is the molarity of a solution prepared by dissolving 15.0 g of NaOH in enough water to make a total of 225 mL of solution?

Solution:

1 mol of NaOH has a mass of 40.00 g, so

#"Moles of NaOH" = 15.0 cancel("g NaOH") × "1 mol NaOH"/(40.00 cancel("g NaOH")) = "0.375 mol NaOH"#

#"Litres of solution" = 225 cancel("mL soln") × "1 L soln"/(1000 cancel("mL soln")) = "0.225 L soln"#

#"Molarity" = "moles of solute"/"litres of solution" = "0.375 mol"/"0.225 L" = "1.67 mol/L"#

Some students prefer to use a "molarity triangle".

It summarizes the molarity formulas as

#"Moles" = "molarity × litres"#

#"Molarity" = "moles"/"litres"#

#"Litres" = "moles"/"molarity"#