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

Go to Problems #11 - 20

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The key point in the problems below will be the molar ratio between acid and base. Problems 1, 2 and 3 utilize a 1:1 ratio. Problems 4 thru 7 use a 2:1 ratio, problem 8 uses a 3:1 ratio and problems 9 and 10 use a 3:2 ratio.

A word to the wise: often 1:1 ratios are taught, but 2:1 ratios are tested. The 2:1 ratios probably will not show up in lecture, but will be included in your homework.


Problem #1: If 20.60 mL of 0.0100 M aqueous HCl is required to titrate 30.00 mL of an aqueous solution of NaOH to the equivalence point, what is the molarity of the NaOH solution?

Solution #1 (the general solution):

1) Write the chemical equation for the reaction:

HCl + NaOH ---> NaCl + H2O

2) The key molar ratio . . . :

. . . is that of HCl to NaOH, a 1 to 1 ratio.

3) Determine moles of HCl:

moles = MV = (0.0100 mol/L) (0.02060 L) = 0.000206 mol

4) Determine moles of NaOH:

1 is to 1 as 0.000206 mol is to x

x = 0.000206 mol of NaOH consumed

5) Determine molarity of NaOH solution:

0.000206 mol / 0.03000 L = 0.00687 M

Solution #2 (specific to a 1:1 ratio):

M1V1 = M2V2

(0.0100 mol/L) (20.60 mL) = (x) (30.00 mL)

x = 0.00687 M


Problem #2: How many milliliters of 0.105 M HCl are needed to titrate each of the following solutions to the equivalence point:

a) 22.5 mL of 0.118 M NH3
b) 125.0 mL of a solution that contains 1.35 of NaOH per liter

We will ignore the fact that HCl-NH3 is actually a strong-weak titration. We are only interested in the volume required for the equivalence point, not the pH at the equivalence point.

Solution to a (using the general solution technique):

1) Chemical equation:

HCl + NH3 ---> NH4Cl

2) HCl to NH3 molar ratio:

1 : 1

3) Moles NH3:

moles = MV = (0.118 mol/L) (0.0225 L) = 0.002655 mol

4) Determine moles of HCl used:

1 is to 1 as x is to 0.002655 mol

x = 0.002655 mol of HCl

5) Determine volume of HCl:

0.105 mol/L = 0.002655 mol / x

x = 0.0253 L = 25.3 mL (to three sig figs)

Solution to b (using the solution technique specific to 1:1 ratios):

1) Determine molarity of NaOH solution:

MV = mass / molar mass

(x) (1.00 L) = 1.35 g / 40.00 g/mol

x = 0.03375 M (I'll round off the final answer to the problem.)

2) Write the chemical equation and state the key molar ratio:

HCl + NaOH ---> NaCl + H2O

1:1 (the HCl to NaOH ratio)

3) Determine volume of HCl required:

M1V1 = M2V2

(0.03375 mol/L) (125.0 mL) = (0.105 mol/L) (x)

x = 40.18 mL (to three sig figs, this is 40.2 mL)


Does the molar ratio (remember, the ChemTeam maintains this is the key insight) always come from the coeffcients of the balanced chemical equation?

Yes.

Does the solution specific to the 1: 1 ratio ever work with other ratios?

No.


Problem #3: How many milliliters of 0.116 M H2SO4 will be needed to titrate 25.0 mL of 0.00840 Ba(OH)2 to the equivalence point:

Solution:

1) Chemical equation:

H2SO4 + Ba(OH)2 ---> BaSO4 + 2H2O

2) Molar ratio:

1 : 1

3) Let's use the specific solution:

M1V1 = M2V2

(0.116 mol/L) (x) = (0.00840 mol/L) (25.0 mL)

x = 1.81 mL (to three sig figs)


The reason I wrote the problem just above is because H2SO4 and Ba(OH)2 show up often in problems where the ratio is not 1:1, examples of which you will see just below. I did not want you to gain the impression that chemicals such as H2SO4 and Ba(OH)2 can NEVER be involved in a 1:1 ratio.


Problem #4: 27.0 mL of 0.310 M NaOH is titrated with 0.740 M H2SO4. How many mL of H2SO4 are needed to reach the end point?

Solution:

1) Moles NaOH present:

(0.310 mol/L) (0.027 L) = 0.00837 mol

2) NaOH to H2SO4 molar ratio is . . . :

. . . 2 : 1

This can be seen from the balanced chemical equation:

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

3) So:

2 is to 1 as 0.00837 mol is to x

0.00837 mol divided by 2 = 0.004185 mol of H2SO4 required

4) Calculate volume of H2SO4 required:

0.004185 mol divided by 0.740 mol/L = 0.0056554 L

5.66 mL (to three sig figs)


I answered this Yahoo Answers question:

"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?"

I think it's a good question and I wanted to show you the question first, if you wanted to ponder it before looking at my answer.

Here is my answer.


Problem #5: 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

(a) What is the equation of the reaction?
(b) What was the molarity of calcium hydroxide solution?

Solution:

1) Chemical equation:

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

2) Rest of solution to problem:

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 #6: 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.

(The ChemTeam copied this solution from an Internet cavewall because he thinks it's a good solution. Notice the use of millimoles rather than moles.)

Solution:

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

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

n = 16.0 mol/L 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 moles/L = 640 mL


Problem #7: If 0.2501 grams of dry sodium carbonate requires 27.00 mL of HCl for complete reaction, what is the molar concentration of HCl?

Solution:

Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O

moles Na2CO3 ---> 0.2501 g / 105.988 g/mol = 0.0023597 mol

2 moles of HCl are required for every mole of Na2CO3

0.0023597 mol times 2 = 0.0047194 mol of HCl

0.0047194 mol / 0.02700 L = 0.1748 M


Problem #8: 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

The key is the 1 : 3 molar ratio between H3Cit and NaOH

moles NaOH ---> (0.0148 mol/L) (0.03227 L) = 0.000477596 mol

1 is to 3 as x is to 0.000477596 mol

x = 0.0001592 mol (of H3Cit)

0.0001592 mol / 0.0250 L = 0.00637 M (to three sig figs)


Problem #9: 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:

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


Problem #10: 51.00 ml of phosphoric acid solution (H3PO4) reacts with 13.90 grams of barium hydroxide, Ba(OH)2, according to the following balanced equation. What is the molarity of the phosphoric acid?

Solution:

2H3PO4 + 3Ba(OH)2 ---> Ba3(PO4)2 + 6H2O

moles Ba(OH)2: 13.90 g / 171.344 g/mol = 0.08112335 mol

3 moles Ba(OH)2 react with 2 moles H3PO4

0.08112335 mol Ba(OH)2 react with x mol H3PO4

x = 0.0540822 mol

molarity of phosphoric acid: 0.0540822 mol/0.05100 L = 1.06 M


Go to Problems #11 - 20

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