Titration to the equivalence point: Determine unknown molarity (or volume) when a strong acid (base) is titrated with a strong base (acid)
Problems #11 - 20

The ten examples

Problems #1-10

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Problem #11: 60.0 mL of 0.10 M Ca(OH)2 solution is mixed with 50.0 mL of 0.10 M HNO3. What is the volume of 0.10 M H2SO4 needed to neutralize the mixture?

Solution #1:

moles Ca(OH)2 ---> (0.10 mol/L) (0.060 mL) = 0.0060 mol

moles HNO3 ---> (0.10 mol/L) (0.050 mL) = 0.0050 mol

Ca(OH)2 + 2HNO3 ---> Ca(NO3)2 + 2H2O

Two HNO3 are required to neutralize one Ca(OH)2, so only 0.0025 mol of Ca(OH)2 will be neutralized.

0.0060 mol minus 0.0025 mol = 0.0035 mol of Ca(OH)2 remaining

Ca(OH)2 and H2SO4 react in a 1 to 1 molar ratio, so 0.0035 mol of H2SO4 is required.

volume H2SO4 ---> 0.0035 mol / 0.10 mol/L = 0.035 L = 35.0 mL

Solution #2:

Write a balanced equation for the first step:

Ca(OH)2 + 2HNO3 ---> Ca(NO3)2 + 2H2O

You need 2 moles of HNO3 for every mole of Ca(OH)2. How much was provided?

n = (60 mL) (0.10 M) = (0.060 L) (0.10 mol/L) = 0.006 mol of Ca(OH)2
n = (50 mL) (0.10 M) = (0.050 L) (0.10 mol/L) = 0.005 mol of HNO3

If we react these fully, 0.005 moles of HNO3 will react with 0.0025 moles of Ca(OH)2. This leaves 0.006 - 0.0025 = 0.0035 moles of Ca(OH)2. Note that we are left with aqueous Ca(OH)2 and those extra hydroxide ions will make the solution basic, so it does make sense that we need to add more acid to neutralize it. Now we write an equation for the next reaction:

Ca(OH)2 + H2SO4 ---> CaSO4 + 2H2O

We need one mole of H2SO4 for every one mole of Ca(OH)2. That makes it easy. We have 0.0035 moles of Ca(OH)2 left over, so we need to add 0.0035 moles of H2SO4.

To get volume, we need to divide the number of moles by the molarity:

V = n / M = (0.0035 mol) / (0.10 M) = 0.035 L = 35 mL of H2SO4

Problem #12: 50.0 mL of 2% (w/w) solution of NaOH is neutralized completely with 0.50 M H2SO4. Calculate (a) molality, (b) molarity of the NaOH solution given the density is 1.08 g/mL, and (c) the volume of 0.50 M H2SO4

Solution:

2% (w/w) means 2 g of NaOH are present in every 100 g of solution.

Since we have 50 mL of solution, we need to know its mass:

50.0 mL times 1.08 g/mL = 54.0 g

grams of NaOH in the 54 g of solution:

2 is to 100 as x is to 54

x = 1.08 g

moles of NaOH ---> 1.08 g / 40.0 g/mol = 0.027 mol

molarity ---> 0.027 mol / 0.050 L = 0.54 M

For molality, we need the mass of water:

54.0 g minus 1.08 g = 52.92 g

molality ---> 0.027 mol / 0.05292 kg = 0.51

For part (c):

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

We know that 2 NaOH are required for every one H2SO4, so do this:

0.027 mol / 2 = 0.0135 mol (this is how much H2SO4 is neutralized by our 0.027 mol of NaOH)

volume = 0.0135 mol / 0.50 mol/L = 0.027 L = 27 mL


Problem #13: 2.00 liters of NH3 at 30.0 °C and 0.200 atm is neutralized by 134 ml of H2SO4 solution. Calculate the molarity of the acid solution.

Solution:

1) We need to get the moles of NH3:

PV = nRT

(0.200 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (303 K)

n = 0.016087 mol

2) Now, the chemical equation for the neutralization:

2NH3 + H2SO4 ---> (NH4)2SO4

The key is that 2 NH3 are required for every one H2SO4.

0.016087 mol / 2 = 0.0080435 mol <--- that's how many moles of H2SO4 reacted

3) The molarity:

0.0080435 mol / 0.134 L = 0.0600 M

Problem #14: 25.0 mL NaHCO3 containing 4.20 g/L required 12.5 mL of HCl to completely react with it. Calculate the molarity of the HCl solution.

Solution:

1) We need to know how many moles of NaHCO3 are present. To do that, we first get grams:

4.2 g is to 1000 mL as x is to 25.0 mL

x = 0.105 g

2) Then, moles:

0.105 g / 84.0059 g/mol = 0.0012499 mol

3) From the balanced equation:

NaHCO3 + HCl = NaCl + H2O + CO2

we see that the molar ratio between NaHCO3 and HCl is one to one. This means that 0.0012499 mol of HCl reacted.

4) The molarity of the HCl:

0.0012499 mol / 0.0125 L = 0.100 M (to three sig figs)

Problem #15: 25.0 mL of a solution of an acid HxA containing 0.10 mol of the acid in each 1000 mL of solution reacts with 75.0 mL of a solution of 0.10 M NaOH. What is the value of x?

Solution:

moles of NaOH ---> (0.10 mol/L) (0.0750 L) = 0.00750 mol

moles of H+ (from HxA) must equal moles of OH-

moles HxA ---> (0.10 mol/L) (0.0250 L) = 0.0025 mol

x = 0.00750 / 0.00250 = 3


Problem #16: In the reaction between an acid HyA and 0.100 mol dm-3 NaOH solution, 25.0 cm3 of a solution of 0.100 mol dm-3 HyA reacts with 50.0 cm3 of the 0.100 mol dm-3 NaOH. What is the value of y?

Solution:

moles of NaOH ---> (0.100 mol/L) (0.0500 L) = 0.00500 mol

moles of H+ (from HyA) must equal moles of OH-

moles HyA ---> (0.10 mol/L) (0.0250 L) = 0.0025 mol

y = 0.00500 / 0.00250 = 2


Problem #17: (a) You titrate a 5.00 mL sample of a generic monoprotic acid, HA, with 27.0 mL of a 0.160 M sodium hydroxide solution. Compute the molarity of the acid to the appropriate number of significant figures.

(b) If the unknown acid had been H2SO4 with the same concentration as HA (calculated above), how much NaOH would have been required for the titration?

Solution:

For (a), this is the reaction:
HA + NaOH ---> NaA + H2O

The key is that there is a 1:1 molar ratio between HA and NaOH.

moles NaOH ---> (0.160 mol/L) (0.02700 L) = 0.00432 mol

From the 1:1 ratio, we know that 0.00432 mol of HA got neutralized.

molarity of HA ---> 0.00432 mol / 0.00500 L = 0.864 M

For (b), write the chemical equation:

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

Notice that, for every ONE H2SO4 neutralized, TWO NaOH are required.

The same molarity of H2SO4 (the 0.864 value) would required TWICE as much NaOH as compared to neutralizing 0.864 M HA.

54.0 mL is the answer.


Problem #18: You dilute 20.0 ml L of a stock solution of H2SO4 to 1.00 L. You then make duplicate titrations of 20.0 mL each of the diluted H2SO4 solution with your 0.202 M NaOH solution to the end point. Titration 1 requires 39.05 mL; titration 2 requires 39.09 mL. Calculate the concentration of the stock H2SO4 solution.

Solution:

1) Use the average of the two volumes of NaOH:

(39.05 mL + 39.09 mL) / 2 = 39.07 mL

2) Moles of NaOH used:

(0.202 mol/L) (0.03907 L) = 0.00789214 mol

3) The balaced chemical equation is:

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

Two NaOH are used for every one H2SO4 neutralized.

4) Moles of H2SO4 neutralized:

0.00789214 mol / 2 = 0.00394607 mol

5) The amount of H2SO4 just above was in 20.0 mL and this volume came from the 1.00 L. How many moles of H2SO4 are present in the 1.00 L?

0.00394607 mol is to 0.0200 L as x is to 1.00 L

x = 0.1973035 mol

6) All the above moles of H2SO4 was originally contained in the 20.0 mL that was then diluted to 1.00 L. What is the molarity of the original 20.0 mL of stock solution?

0.1973035 mol / 0.0200 L = 9.865175 M

To three sig figs, 9.86 M


Problem #19: 12.0 mL of 0.100 M NaOH is mixed with 48.0 mL of Ba(OH)2 of unknown concentration. The solution is completely neutralized by 47.7 mL of 0.0500 M H2SO4. What is the concentration of the original Ba(OH)2 solution?

Solution:

1) We will first ignore the presence of the Ba(OH)2 solution and determine how much H2SO4 was neutralized by the NaOH.

moles NaOH ---> (0.1 mol/L) (0.012 L) = 0.0012 mol

Two moles of NaOH are required for every one mole of H2SO4.

moles H2SO4 neutralized ---> 0.0012 mol / 2 = 0.0006 mol

volume H2SO4 used ---> 0.0006 mol / 0.05 mol/L = 0.012 L

2) The rest of the volume of H2SO4 was used to neutralize the Ba(OH)2 solution.

47.7 - 12.0 = 35.7 mL

3) We now require the moles of Ba(OH)2. We get this by using the remaining H2SO4.

moles H2SO4 ---> (0.05 mol/L) (0.0357 L) = 0.001785 mol

4) The H2SO4 and Ba(OH)2 react in a 1:1 molar ratio. Therefore:

moles Ba(OH)2 = 0.001785 mol

molarity of Ba(OH)2 ---> 0.001785 mol / 0.048 L = 0.0372 M


Problem #20: A 20.0 mL solution of triprotic acid is titrated to neutralization by 42.0 mL of 0.0483 M KOH solution. What is the molarity of the acid solution?

Solution:

1) Call the triprotic acid H3A. Its reaction with KOH is this:

H3A + 3KOH ---> K3A + 3H2O

We see three moles of KOH are required for each mole of H3A that reacts.

2) The relationship between amounts of acid and base that are required for neutralization is codified here:

MaVa   MbVb
–––––  =  –––––
na   nb

3) Insert values in place and solve:

(x) (20.0 mL)   (0.0483 M) (42.0 ml)
–––––––––––  =  ––––––––––––––––
1   3

x = 0.0338 M


Problem #21: A 20.0 mL solution of triprotic acid is titrated to neutralization by 42.0 mL of 0.0483 M Al(OH)3 solution. What is the molarity of the acid solution?

Solution:

1) Call the triprotic acid H3A. Its reaction with Al(OH)3 is this:

H3A + Al(OH)3 ---> AlA + 3H2O

We see one mole of Al(OH)3 is required for each mole of H3A that reacts.

2) The relationship between amounts of acid and base that are required for neutralization is codified here:

MaVa   MbVb
–––––  =  –––––
na   nb

3) Insert values in place and solve:

(x) (20.0 mL)   (0.0483 M) (42.0 ml)
–––––––––––  =  –––––––––––––––––
1   1

x = 0.101 M (to three sig figs)


The ten examples

Problems #1-10

Return to the Acid Base menu

Return to a listing of many types of acid base problems and their solutions