__Neutralization
Reactions – Finding unknown M or unknown V__** Std 5**

All neutralization reactions can be handled with this very
simple method. It is a slight variation
on the M_{1}V_{1}
= M_{2}V_{2} equation.
But first let’s be sure we know what is happening in **neutralization**.

**H ^{+} + OH^{-} ----->
H_{2}O**

Three things can happen when
you mix acidic and basic solutions:

a. You wind
up with **more H ^{+} ions** than

b. You wind
up with **more ** than H

c. You wind up with equal numbers of H^{+} and ^{-}**neutral**. We call this **neutralization**.

If you have a 1 M HCl
solution, each HCl molecule will give you 1 H^{+} ion. This means the molarity of the H^{+}
ions will also be 1 M.

It is different for H_{2}SO_{4
} because each H_{2}SO_{4}
molecule produces 2 H^{+ }ions.
So, a 1 M H_{2}SO_{4 } solution will create 2 H^{+} ions for
every H_{2}SO_{4}^{ } molecule.
This makes the molarity of the H^{+} ions 2 M.

The same thing happens with bases like Ba(OH)_{2
} that happens with H_{2}SO_{4}
, except that Ba(OH)_{2} is producing ^{-}^{+}
ions. So, a 1 M Ba(OH)_{2} will produce 2 ^{-}_{2 } . This will create a 2 M OH^{-} ion
solution.

When you
are given molarities, ** this will be the molarity of the molecule of H_{2}SO_{4}
or Ba(OH)_{2} , not the H^{+} or **.

Using
a small change to the M_{1}V_{1} = M_{2}V_{2} equation will simplify your
calculations. In neutralization
reactions we have to account for the H^{+} and ^{-}

M |

a = the # H^{+} ions in the acid molecule and
M_{1}V_{1} is for the acid

b = ^{-}_{2}V_{2}
is for a base

**Examples:**

1. Given 10 mL of a 1 M HCl, what is the molarity of a NaOH
solution if 10 mL neutralizes the HCl solution?

(1 M_{ HCl} )(10 mL)(1) = (? M_{
NaOH} )(10mL)(1)

**Ans:** 1 M = M_{ NaOH}

**ANALYSIS OF ANSWER**: This
problem didn’t require the new formula because HCl gives only 1 H^{+}
and NaOH also gives just 1 OH^{-} ion.
So the formula works as is because a & b are both = 1.

2. Given 10 mL of a 1 M H_{2}SO_{4} , what is
the molarity of a NaOH solution if 10 mL will neutralize the H_{2}SO_{4} solution?

(1 M H_{2}SO_{4} )(10
mL)(2) =
(? M_{ NaOH} )(10mL)(1)

**Ans:** 2 M = M_{ NaOH}

**ANALYSIS OF ANSWER**: This
problem did require the new formula. H_{2}SO_{4}
gives 2 H^{+} ions, which makes a = 2. NaOH just gives 1 ^{-}^{ } ion, so b = 1 .

3. Given 10 mL of a 1 M HCl , what is the molarity of a Ba(OH)_{2}
solution if 10 mL neutralizes the HCl solution?

(1 M HCl )(10 mL)(1) = ( ?
M Ba(OH)_{2} )(10mL)(2)

**Ans:** ½ M = M
Ba(OH)_{2}

**ANALYSIS OF ANSWER**: This
problem did require the new formula because each Ba(OH)_{2} molecule
gives off 2 ^{-}^{
} ions, making b = 2. Since we have equal amounts of the two
solutions, the Ba(OH)_{2} only needs to be half as strong as the HCl
solution, where a = 1.

4. Given 10 mL of a 1 M H_{3}PO_{4} , what is
the molarity of a Ba(OH)_{2} solution if 10 mL neutralizes the HCl
solution?

(1 M H_{3}PO_{4} )(10
mL)(3) =
( ? M Ba(OH)_{2} )(10mL)(2)

**Ans:** 1.5 M = M
Ba(OH)_{2}

**ANALYSIS OF ANSWER**: This
problem did require the new formula because each Ba(OH)_{2} molecule
gives off 2 ^{-}^{
} ions (making b = 2), and each H_{3}PO_{4
} molecule gives off 3 H^{+}
ions (making a = 3) . Since we have equal
amounts of the two solutions, the Ba(OH)_{2} only needs to be a little
stronger to balance with 3 H^{+} ions given off the the H_{3}PO_{4}^{
} molecules.

**The next problem demonstrates that you can use this
formula for unknown volume problems in the same way we use it for unknown
molarity problems.**

5. Given 10 mL of a 1 M H_{3}PO_{4} , and a 1 M
NaOH solution, how much NaOH is needed to neutralize the H_{3}PO_{4}
solution?

(1 M H_{3}PO_{4} )(10
mL)(3) =
( 1 M NaOH)( V )(1)

**Ans:** 30 mL =
V (of NaOH)

**ANALYSIS OF ANSWER**: This
problem did require the new formula because each Ba(OH)_{2} molecule
gives off 2 ^{-}^{
} ions (making b = 2), and each H_{3}PO_{4
} molecule gives off 3 H^{+}
ions (makina a = 3). Since we have equal
amounts of the two solutions, the Ba(OH)_{2} only needs to be a little
stronger to balance with 3 H^{+} ions given off the the H_{3}PO_{4}^{
} molecules.