Ten Examples

The first five examples below use a 1:1 molar ratio to solve the problem. In the next five examples (that is, examples 6 to 10), there are some 2:1 ratios and a 3:2 ratio. The additional problems in the linked files contain a number of 2:1 molar ratio problems, several 1:1 ratios and one 3:2 ratio.

**Example #1:** How many milliliters of 0.122 M HCl would be required to titrate 6.45 g KOH?

**Solution #1:**

1) The balanced equation for the reaction:

HCl(aq) + KOH(s) ---> KCl(aq) + H_{2}O(ℓ)

2) The HCl and the KOH react in a 1:1 molar ratio. This means:

moles HCl used = moles KOH used

3) Consequently, we can use:

MV = mass / molar mass(0.122 mol/L) (x) = 6.45 g / 56.1049 g/mol

x = 0.9423215 L

to three sig figs, 942 mL

However, using the above only works when there is a 1:1 molar ratio between reactants. Below is the more general solution.

**Solution #2:**

1) moles KOH:

6.45 g / 56.1049 g/mol = 0.114963 mol

2) molar ratio:

From the balanced equation, we see that the HCl:KOH molar ratio is 1:1. Therefore:0.114963 mol of HCl was used

3) volume of HCl:

0.114963 mol / 0.122 mol/L = 0.94232 Lto three sig figs, 942 mL

**Example #2:** What is the molarity of a hydrochloric acid solution if 25.0 mL of the solution reacts completely with 1.66 g NaHCO_{3}?

**Solution:**

1) The balanced equation for the reaction:

HCl(aq) + NaHCO_{3}(s) ---> NaCl(aq) + CO_{2}(g) + H_{2}O(ℓ)

2) Note the 1 to 1 molar relationship between HCl and NaHCO_{3}.

moles NaHCO_{3}---> 1.66 g / 84.0059 g/mol = 0.01976 mol

3) The 1:1 ratio means:

0.01976 mol of HCl reacted with 0.01976 mol of NaHCO_{3}0.01976 mol / 0.0250 L = 0.790 M (to three sig figs)

**Example #3:** In the following acid-base neutralization, 1.68 g of the solid acid phenol (HC_{6}H_{5}O; MW = 94.12 g/mol) neutralized 11.61 mL of aqueous NaOH solution. Calculate the molarity of the base solution.

**Solution:**

1) The chemical reaction is this:

HC_{6}H_{5}O(aq) + NaOH(aq) ---> NaC_{6}H_{5}O(aq) + H_{2}O(ℓ)The key is that there is a one-to-one molar ratio between the phenol (formula also seen as C

_{6}H_{5}OH. However, it's an acid) and the sodium hydroxide.

2) Moles phenol:

1.68 g / 94.12 g/mol = 0.01785 mol

3) Molarity of NaOH solution:

Due to the one-to-one molar ratio, 0.01785 mol of NaOH reacted.0.01785 mol / 0.01161 L = 1.54 M

**Example #4:** How many mL of 0.258 M NaOH are required to completely neutralize 2.00 g of acetic acid (HC_{2}H_{3}O_{2})?

**Solution:**

1) The chemical reaction is this:

HC_{2}H_{3}O_{2}(aq) + NaOH(aq) ---> NaC_{2}H_{3}O_{2}(aq) + H_{2}O(ℓ)The key is that there is a one-to-one molar ratio between the acetic acid and the sodium hydroxide.

2) Moles acetic acid:

2.00 g / 60.0516 g/mol = 0.0333047 mol

3) Volume of NaOH solution:

Due to the one-to-one molar ratio, 0.0333047 mol of NaOH reacted.0.0333047 mol / 0.258 mol/L = 0.129 L = 129 mL

Comment about Examples #3 and #4: both phenol and acetic acid are weak acids. However, since we are not concerned with the pH of the solution, there is no need to distinguish between strong and weak. In other words, it takes exactly the same amount of base to neutralize a given amount of acid and it does not matter if the acid is strong or weak.

**Example #5:** When 5.231 g of calcium hydroxide is reacted with a 29.0 mL of a 0.100 M sulfuric acid solution, what volume of the H_{2}SO_{4} solution is required for complete neutralization?

**Solution:**

1) The chemical reaction is this:

H_{2}SO_{4}(aq) + Ca(OH)_{2}(aq) ---> CaSO_{4}(s) + 2H_{2}O(ℓ)The key is that there is a one-to-one molar ratio between the sulfuric acid and the calcium hydroxide.

2) Moles calcium hydroxide:

5.231 g / 74.0918 g/mol = 0.0706016 mol

3) Volume of H_{2}SO_{4} solution:

Due to the one-to-one molar ratio, 0.0706016 mol of H_{2}SO_{4}reacted.0.0706016 mol / 0.100 mol/L = 0.706 L = 706 mL

Notice that both calcium hydroxide and sulfuric acid often show up in problems that use a 2:1 molar ratio. However, when paired together, a 1:1 ratio is used. The same can be said for a problem involving H_{3}PO_{4} and Al(OH)_{3}. A 1:1 molar ratio would be used. See problem #9 in the first file of additional problems.

**Example #6:** How many mL of 2.5 M H_{2}SO_{4} are needed to neutralize 75.0 grams of NaOH?

**Solution:**

H_{2}SO_{4}+ 2NaOH ---> Na_{2}SO_{4}+ 2H_{2}O75.0 g / 40.0 g/mol = 1.875 mol

Based on the 1:2 molar ratio between H

_{2}SO_{4}and NaOH, we see that it takes two NaOH for every one H_{2}SO_{4}1.875 mol / 2 = 0.9375 mol of H

_{2}SO_{4}neutralized0.9375 mol / 2.5 mol/L = 0.375 L = 375 mL

**Example #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:**

Na_{2}CO_{3}+ 2HCl ---> 2NaCl + CO_{2}+ H_{2}Omoles Na

_{2}CO_{3}---> 0.2501 g / 105.988 g/mol = 0.0023597 mol2 moles of HCl are required for every mole of Na

_{2}CO_{3}(0.0023597 mol) (2) = 0.0047194 mol of HCl

0.0047194 mol / 0.02700 L = 0.1748 M

**Example #8:** A 250.0 cm^{3} solution of NaOH was prepared. 25.0 cm^{3} was used as a sample. The 25.0 cm^{3} solution required 28.2 cm^{3} of 0.100 mol dm¯^{3} of HCl for neutralization. Calculate what mass of NaOH was dissolved to make up the original 250.0 cm^{3} solution.

**Solution:**

1) Important starting information:

HCl + NaOH ---> NaCl + H_{2}O

Molar ratio between HCl and NaOH is 1 : 1

2) Moles HCl needed to neutralize the NaOH in the sample:

(0.100 mol dm¯^{3}) (0.0282 dm^{3}) = 0.00282 molNote that the change from cm

^{3}to dm^{3}is the same as the change from mL to L. There are 1000 cm^{3}in 1 dm^{3}.In other words, 1 dm

^{3}= 1 L. Therefore, 0.100 mol dm¯^{3}is EXACTLY 0.100 M.

3) Moles of NaOH neutralize, then the NaOH molarity:

Based on the 1:1 molar ratio, we know that 0.00282 mol of HCl neutralized 0.00282 mol of NaOH.0.00282 mol / 0.0250 dm

^{3}= 0.1128 MWe could have used M

_{1}V_{1}= M_{2}V_{2}to get the molarity of the NaOH. I chose to do the longer calculation.

4) Mass of NaOH is the 250.0 cm^{3} solution:

MV = mass / molar mass(0.1128 mol dm¯

^{3}) (0.2500 dm^{3}) = x / 40.0 g/molx = 1.128 g (rounded to three sig figs, it is 1.13 g)

**Example #9:** What mass of calcium carbonate, CaCO_{3} will a 31.27 mL sample of 0.2442 M phosphoric acid, H_{3}PO_{4}, neutralize according to the reaction shown below? The molar mass of calcium carbonate is 100.09 g/mol.

**Solution:**

1) The reaction of interest is this:

3CaCO_{3}(s) + 2H_{3}PO_{4}(aq) ---> Ca_{3}(PO_{4})_{2}(s) + 3CO_{2}(g) + 3H_{2}O(ℓ)

2) Moles of phosphoric acid that react:

moles = MVmoles = (0.2442 mol/L) (0.03127 L) = 0.007636134 mol

3) Determine moles of CaCO_{3} consumed:

The molar ratio between CaCO_{3}and H_{3}PO_{4}is 3:23 is to 2 as x is to 0.007636134 mol

x = 0.011454201 mol of CaCO

_{3}consumed

4) Determine mass of CaCO_{3} consumed:

(0.011454201 mol) (100.09 g/mol) = 1.146 g (to 4 sig figs)

**Example #10:** 1.314 g of sodium carbonate was dissolved and made up to a final volume of 250.0 mL. This solution was used to determine the concentration of a solution of hydrochloric acid. Four 25.0 mL samples of the acid were titrated with the sodium carbonate solution. The average titration volume required to reach the end point was 23.45 mL. Calculate the concentration of the hydrochloric acid solution.

**Solution:**

1) Determine the molarity of the sodium carbonate solution:

MV = mass / molar mass(M) (0.2500 L) = 1.314 g / 105.99 g/mol

molarity = 0.04958958 M

2) Determine the moles of sodium carbonate in the average volume of 23.45 mL:

MV = moles(0.04958958 mol/L) (0.02345 L) = moles

0.001162875651 mol

Note: be careful about which volume goes where. When I first solved this problem, I used 25.0 mL for the sodium carbonate. Oopsie!!

3) Write the chemical equation for the reaction:

Na_{2}CO_{3}+ 2HCl ---> 2NaCl + CO_{2}+ H_{2}OThe key point is the 1:2 molar ratio between sodium carbonate and HCl.

4) Determine moles of HCl consumed, then molarity of HCl solution:

(0.001162875651 mol) (2) = 0.002325751302 mol0.002325751302 mol / 0.02500 L = 0.09303 M (to four sig figs)

**Bonus Example #1:** What is the gram-formula weight of RbOH when 2.050 g is dissolved in 250.0 mL of water. 25.00 mL of the solution is neutralized by 18.18 mL of 0.1100 M HCl? (Gram-formula weight is an older term for what is now simply called molar mass.)

**Solution:**

RbOH + HCl ---> RbCl + H_{2}OThe key point from the equation is that RbOH and HCl react in a 1:1 molar ratio.

Moles HCl that reacted ---> (0.1100 mol/L) (0.01818 L) = 0.0019998 mol

The 25.00 mL of the RbOH solution was neutralized by 0.0019998 mol of HCl. Since the molar ratio is 1:1 between the RbOH and HCl in the balanced reaction, this means that the 25.00 mL of the RbOH solution contained 0.0019998 mol of RbOH.

The original 250.0 mL of RbOH solution has ten times the volume of the 25.0 mL sample, so the 250.0 mL must have contained 0.019998 mol of RbOH (i.e., ten times the number of moles contained in the 25.0 mL sample).

The gram-formula weight of RbOH would then be 2.050 g/0.019998 mol of RbOH = 102.5 g/mol which to four sig figs is the answer you would find if you calculated the gram-formula weight using the periodic table.

**Bonus Example #2:** A vinegar sample is titrated with 0.52 M NaOH. In one trial, 0.49 g of vinegar requires 0.73 g of NaOH to reach the endpoint. Find the percent by mass of acetic acid in this vinegar sample. (Density of vinegar is 1.00 g/mL)

**Solution:**

1) That 0.73 g of NaOH means 0.73 g of NaOH solution. (Solving it with 0.73 g of NaOH gives an impossible answer.) In addition, we would need the density of the NaOH solution so as to determine the volume of solution. 0.52 M solution probably has a density close to 1.00 g/mL, so I'll use that.

0.73 g / 1.00 g/mL = 0.73 mL = 0.00073 L0.52 mol/L times 0.00073 L = 0.0003796 mol of NaOH

This means 0.0003796 mol of acetic acid. (Based on the 1:1 molar ratio between NaOH and acetic acid when they react.)

0.0003796 mol times 60.0516 g/mol = 0.0227956 g

0.0227956 g / 0.49 g = 0.0465 = 4.6%

Note that the density of vinegar is not needed.

2) So, you're curious about that impossible answer?

0.73 g / 39.9969 g/mol = 0.0182514 mol of NaOH0.0182514 mol of NaOH means 0.0182514 mol of acetic acid reacted.

(0.0182514 mol) (60.0516 g/mol) = 1.096 g of acetic acid

1.095 g / 0.49 g = 2.2367 = 223.67%