### Titration to the equivalence point: Determine unknown molarity (or volume) when a strong acid (base) is titrated with a strong base (acid)Problems #1 - 10

Problem #0: When H2SO4 and NaOH is reacted, is it the H+ and OH¯ that has equal number of moles, or the H2SO4 and NaOH has equal number of moles?

Solution:

1) It will be the H+ and the OH¯ that have the equal number of moles.

2) Here is the complete molecular equation:

H2SO4(aq) + 2NaOH(aq) ---> Na2SO4(aq) + 2H2O

This is where the 1:2 molar ratio of H2SO4 and NaOH comes from.

3) Here is the complete ionic equation:

2H+(aq) + SO42¯(aq) + 2Na+(aq) + 2OH¯(aq) ---> 2Na+(aq) + SO42¯(aq) + 2H2O(ℓ)

Notice I am treating the sulfuric acid as if it is 100% ionized in both hydrogens. This is the the true case of how sulfuric acid ionizes in solution, but I am here more interested in the stoichiometry of the reaction between sulfuric acid and sodium hydroxide, not in how sulfuric acid ionizes by itself in solution.

4) Here is the unreduced net-ionic equation:

2H+(aq) + 2OH¯(aq) ---> 2H2O(ℓ)

2H+ from one H2SO4 and 2OH¯ from 2NaOH. The 1:2 ratio up above gets obscured here. That's because the ratio above is based on the full molecular formula, not just the number of ions involved.

5) Here is the reduced net-ionic:

H+(aq) + OH¯(aq) ---> H2O(ℓ)

This shows that, in an acid base reaction, hydrogen ions and hydroxide ions react in equal molar amounts (which is a 1:1 molar ratio). The actual molecules that provide the ions for reaction may, or may not, be in a 1:1 molar ratio

Problem #1: A volume of 20.00 mL of 0.250 M Al(OH)3 neutralizes a 75.00 mL sample of H2SO4 solution. What is the concentration of H2SO4?

Solution #1:

2Al(OH)3 + 3H2SO4 ---> Al2(SO4)3 + 6H2O

Al(OH)3 and H2SO4 are in a 2 : 3 molar ratio.

moles Al(OH)3 ---> (0.250 mol/L) (20.00 mL) = 5.00 millimoles

2 is to 3 as 5.00 is to x

x = 7.50 millimoles

molarity H2SO4 ---> 7.50 millimoles / 75.00 mL = 0.100 M

Solution #2:

 (x) (75.0 mL) (0.250 M) (20.0 mL) ––––––––––– = ––––––––––––––––– 3 2

(3) (0.250 M) (20.0 mL) = (2) (x) (75.0 mL)

x = 0.100 M

Problem #2: A 21.62 mL sample of Ca(OH)2 solution was titrated with 0.2545 M HCl. 45.87 mL of the acid was required to reach the endpoint of the titration. What was the molarity of calcium hydroxide solution?

Solution:

2HCl + Ca(OH)2 ---> CaCl2 + 2H2O

moles HCl ---> (0.2545 mol/L) (0.04587 L) = 0.011674 mol

use the 2 : 1 molar ratio of HCl to Ca(OH)2:

2 is to 1 as 0.011674 mol is to x

x = 0.005837 mol of Ca(OH)2

molarity of Ca(OH)2 ---> 0.005837 mol / 0.02162 L = 0.2700 M (to four sig figs)

Problem #3: Calculate the volume of NaOH necessary to neutralize 50.0 mL of a 16.0 M solution of sulfuric acid. The concentration of the NaOH is 2.50 M.

Solution:

2NaOH + H2SO4 ---> Na2SO4 + 2H2O

Calculate moles of H2SO4 by using n = C x V:

n = 16.0 mmol/mL x 50 mL = 800 millimoles

Now look at the equation. It says that for every mole of H2SO4 you need twice that many moles of NaOH to neutralize all the H2SO4.

So moles of NaOH = 800 x 2 = 1600 millimoles of NaOH needed.

Now that you have moles and a concentration you can find volume:

V = moles / concentration

V = 1600 millimoles / 2.50 mmoles/mL = 640. mL

Note: I copied this answer from an "answers" website. The original writer wrote the molarity as 2.50 moles/L instead of 2.50 mmol/mL. While 2.50 moles/L is exactly the same as 2.50 mmol/mL, the units in the above expression would not match. If you did that on an answer to a test, you might get a deduction for using the wrong units on the molarity. Moles does not cancel with millimoles, although the numerical answer would be correct.

Problem #4: What is the citric acid concentration in a soda if it requires 32.27 mL of 0.0148 M NaOH to titrate 25.00 mL of soda?

Solution:

Citric acid has three acidic hydrogens, so I will use H3Cit for the formula.

H3Cit + 3NaOH ---> Na3Cit + 3H2O

 MaVa MbVb ––––– = ––––– na nb

 (x) (25.0 mL) (0.0148 M) (32.27 mL) ––––––––––– = ––––––––––––––––– 1 3

(3) (x) (25.0 mL) = (1) (0.0148 M) (32.27 mL)

x = 0.00637 M (to three sig figs)

Problem #5: Consider the unbalanced molecular equation in which citric acid, H3C6H5O7, is reacted with sodium hydroxide:

H3C6H5O7(aq) + NaOH(aq) ---> Na3C6H5O7(aq) + H2O(ℓ)

If 62.7 mL of 1.20 M NaOH(aq) is titrated with 32.0 mL of H3C6H5O7(aq) what the molarity of H3C6H5O7(aq)?

Solution:

1) First, let us balance the equation:

H3C6H5O7(aq) + 3NaOH(aq) ---> Na3C6H5O7(aq) + 3H2O(ℓ)

2) The key point is the 1:3 stoichiometric ratio between the citric acid and the NaOH.

moles NaOH ---> (1.20 mol/L) (0.0627 L) = 0.07524 mol

You need three moles of NaOH to neutralize every one mole of citric acid.

moles citric acid ---> 0.07524 mol / 3 = 0.02508 mol

3) Calculate the molarity of the citric acid:
0.02508 mol / 0.0320 L = 0.78375 M

Three sig figs gives 0.784 M for the final answer.

Problem #6: Calculate the molar concentration of phosphoric acid if 90.0 mL are required to neutralize 200. mL of 0.200 M calcium hydroxide solution.

Solution:

1) First, let us balance the equation:

2H3PO4(aq) + 3Ca(OH)2(aq) ---> Ca3(PO4)2(s) + 6H2O(ℓ)

Notice that solid calcium phosphate is formed. This is immaterial to the following calculation.

2) There is a 2:3 molar ratio between H3PO4 and Ca(OH)2.

moles Ca(OH)2 ---> (0.2 mol/L) (0.2 L) = 0.04 mol

2 is to 3 as x is to 0.04

x = 0.02667 mol of H3PO4

3) Calculate the molarity of the H3PO4:

0.02667 mol / 0.09 L = 0.296296 M

Rounded off to three sig figs, 0.296 M is the H3PO4 concentration.

Tetrahydroxyarsoranyl is an actual compound (I think) and its formula can be rendered as H4AsO4. The ChemTeam does not know if those four hydrogens are acidic hydrogens, but he's going to act like they are.

Problem #7: 35.0 mL of a solution of H4AsO4 is titrated with 0.0840 M Ba(OH)2. 41.4 mL of the base is required to reach the endpoint. Calculate the molarity of the H4AsO4 solution.

Solution:

H4AsO4 + 2Ba(OH)2 ---> Ba2AsO4 + 4H2O

 (x) (35.0 mL) (0.0840 M) (41.4 mL) ––––––––––– = ––––––––––––––––– 1 2

(1) (0.0840 M) (41.4 mL) = (2) (x) (35.0 mL)

x = 0.0497 M

Problem #8: 35.0 mL of a solution of H4AsO4 is titrated with 0.0840 M Al(OH)3. 41.4 mL of the base is required to reach the endpoint. Calculate the molarity of the H4AsO4 solution.

Solution:

3H4AsO4 + 4Al(OH)3 ---> Al4(AsO4)3 + 12H2O

 (x) (35.0 mL) (0.0840 M) (41.4 mL) ––––––––––– = ––––––––––––––––– 3 4

(3) (0.0840 M) (41.4 mL) = (4) (x) (35.0 mL)

x = 0.0745 M

Problem #9: A flask containes 30.0 mL of 0.100 M hydrochloric acid and 20.0 mL of hydrobromic acid. If 50.0 mL of 0.200 M sodium hydroxide was required to neutralize the mixture of acids in the flask, what was the concentration of the hydrobromic acid? (See important comment at end of the one equation solution.)

Solution:

Solution using steps:

moles of NaOH ---> (0.200 mol/L) (0.0500 L) = 0.0100 mol

NaOH and HCl react in a 1:1 molar ratio. Let's see how much NaOH is left over after reacting with the HCl.

moles HCl ---> (0.100 mol/L) (0.0300 L) = 0.00300 mol

0.0100 mol − 0.00300 mol = 0.00700 mol <--- left over NaOH

NaOH and HBr react in a 1:1 molar ratio. To achieve neutralization, 0.00700 mol of NaOH reacts with 0.00700 mole of HBr.

molarity ---> 0.00700 mol / 0.0200 L = 0.350 M

Solution using one equation:

Ma1Va1 / na1 + Ma2Va2 / na2 = MbVb / nb

[(0.100 mol/L) (0.0300 L)] / 1 + [(x) (0.0200 L)] / 1 = [(0.200 mol/L) (0.0500 L)] / 1

0.003 + 0.02x = 0.01

0.02x = 0.007

x = 0.350 M

A colleague has studied the above one equation solution and has prepared some discussion of it. Here are his words to me concerning the one equation solution:

"I find the above a little misleading since it is not generally correct. That is, your first equation above seems to imply that we just divide moles by the stoichiometric coefficients and then just add everything together for the case of two acids and one base. It turns out that numerically this is true only because the stoichiometric coefficient of the strong base (your nb above) is the same in both of the neutralization reactions and equals 1. This is not the case in problem #10 and no such “solution using one equation” can be given as a result.

I wrote up a detailed “Deep Dive” on this one-step method approach. It contrasts the two problems and shows that for two strong acids and one strong base there is no single equation that one can use."

Problem #10: A flask contains 30.0 mL of 0.10 M hydrochloric and 20.0 mL of sulfuric acid. If 50.0 mL of 0.20 M sodium hydroxide was required to neutralize the mixture of acids in the flask, what was the concentration of the sulfuric acid?

Solution:

moles of NaOH ---> (0.20 mol/L) (0.0500 L) = 0.0100 mol

NaOH and HCl react in a 1:1 molar ratio. Let's see how much NaOH is left over after reacting with the HCl.

moles HCl ---> (0.10 mol/L) (0.0300 L) = 0.00300 mol

0.0100 mol − 0.00300 mol = 0.00700 mol <--- left over NaOH

NaOH and H2SO4 react in a 2:1 molar ratio. To achieve neutralization, 0.00700 mol of NaOH reacts with 0.00350 mole of H2SO4.

molarity ---> 0.00350 mol / 0.0200 L = 0.175 M

Bonus Problem: A mixture of hydrochloric acid and sulfuric acid is prepared so that it contains 0.635 M HCl and 0.135 M H2SO4. What volume of 0.140 M NaOH would be required to completely neutralize all of the acid in 588.8 mL of this solution?

Solution:

1) Determine the NaOH volume needed for the HCl:

HCl + NaOH ---> NaCl + H2O

(0.635 mol/L) (0.5888 L) = 0.373888 mol HCl present

By the above chemical equation, HCl and NaOH react in a 1:1 molar ratio. Therefore, 0.373888 mol of NaOH is required.

0.373888 mol / 0.140 mol/L = 2.67063 L

2) Determine the NaOH volume needed for the H2SO4:

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

(0.135 mol/L) (0.5888 L) = 0.079488 mol H2SO4 present

The molar ratio between H2SO4 and NaOH is 1:2. Two moles of NaOH are required for every one mole of H2SO4

(0.079488 mol) (2) = 0.158976 mol NaOH required

0.158976 mol / 0.140 mol/L = 1.13554 L

3) Add the two volumes:

2.67063 L + 1.13554 L = 3.80617 L

To three sig figs, the answer is 3.81 L.