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 M1V1  =  M2V2  equation.  But first let’s be sure we know what is happening in neutralization.

H+  +  OH-  ----->   H2O

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

a.      You wind up with more H+ ions than OH- ions.  These solutions are acid.

b.      You wind up with more OH- ions than H+ ions.   These solutions are basic.

c.      You wind up with equal numbers of H+ and OH- ions.  These solutions are 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 H2SO4  because each H2SO4 molecule produces 2 H+ ions.  So, a 1 M H2SO4  solution will create 2 H+ ions for every H2SO4  molecule.  This makes the molarity of the H+ ions 2 M.

The same thing  happens with bases like  Ba(OH)2   that happens with H2SO4 , except that Ba(OH)2 is producing OH- ions instead of H+ ions.  So, a 1 M Ba(OH)2  will produce 2 OH- ions for every molecule of Ba(OH)2  .   This will create a 2 M OH- ion solution.

When you are given molarities, this will be the molarity of the molecule of H2SO4 or Ba(OH)2 , not the H+ or OH- molecules.

Using a small change to the  M1V1  =  M2V2  equation will simplify your calculations.  In neutralization reactions we have to account for the H+ and OH- ions given off by the molecules that are being mixed.  So we modify the equation as follows:

 M1V1 a =  M2V2 b

a = the # H+ ions in the acid molecule and M1V1 is for the acid

b = OH- ions in the base molecule and M2V2  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 H2SO4 , what is the molarity of a NaOH solution if 10 mL will neutralize the  H2SO4  solution?

(1 M H2SO4 )(10 mL)(2)  =  (? M NaOH )(10mL)(1)

Ans:    2 M =  M NaOH

ANALYSIS OF ANSWER:  This problem did require the new formula.  H2SO4 gives 2 H+ ions, which makes a = 2.    NaOH just gives 1 OH-  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 OH-  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 H3PO4 , what is the molarity of a Ba(OH)2 solution if 10 mL neutralizes the HCl solution?

(1 M H3PO4 )(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 OH-  ions (making b = 2), and each H3PO4  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 H3PO4  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 H3PO4 , and a 1 M NaOH solution, how much NaOH is needed to neutralize the H3PO4 solution?

(1 M H3PO4 )(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 OH-  ions (making b = 2), and each H3PO4  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 H3PO4  molecules.