Problems #1 - 10

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

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 + H_{2}O

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 xx = 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):**

M_{1}V_{1}= M_{2}V_{2}(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 NH_{3}

b) 125.0 mL of a solution that contains 1.35 of NaOH per liter

We will ignore the fact that HCl-NH_{3} 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 + NH_{3}---> NH_{4}Cl

2) HCl to NH_{3} molar ratio:

1 : 1

3) Moles NH_{3}:

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 molx = 0.002655 mol of HCl

5) Determine volume of HCl:

0.105 mol/L = 0.002655 mol / xx = 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 + H_{2}O1:1 (the HCl to NaOH ratio)

3) Determine volume of HCl required:

M_{1}V_{1}= M_{2}V_{2}(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 H_{2}SO_{4} will be needed to titrate 25.0 mL of 0.00840 Ba(OH)_{2} to the equivalence point:

**Solution:**

1) Chemical equation:

H_{2}SO_{4}+ Ba(OH)_{2}---> BaSO_{4}+ 2H_{2}O

2) Molar ratio:

1 : 1

3) Let's use the specific solution:

M_{1}V_{1}= M_{2}V_{2}(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 H_{2}SO_{4} 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 H_{2}SO_{4} 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 H_{2}SO_{4}. How many mL of H_{2}SO_{4} 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 H_{2}SO_{4} molar ratio is . . . :

. . . 2 : 1This can be seen from the balanced chemical equation:

2NaOH + H

_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}O

3) So:

2 is to 1 as 0.00837 mol is to x0.00837 mol divided by 2 = 0.004185 mol of H

_{2}SO_{4}required

4) Calculate volume of H_{2}SO_{4} required:

0.004185 mol divided by 0.740 mol/L = 0.0056554 L5.66 mL (to three sig figs)

I answered this Yahoo Answers question:

"When H_{2}SO_{4}and NaOH is reacted, is it the H^{+}and OH¯ that has equal number of moles, or the H_{2}SO_{4}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.

**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}---> CaCl_{2}+ 2H_{2}O

2) Rest of solution to problem:

moles HCl ---> (0.2545 mol/L) (0.04587 L) = 0.011674 moluse the 2 : 1 molar ratio of HCl to Ca(OH)

_{2}:2 is to 1 as 0.011674 mol is to xx = 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 + H_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}OCalculate moles of H

_{2}SO_{4}by using n = C x V:n = 16.0 mol/L x 50 mL = 800 millimolesNow look at the equation. It says that for every mole of H

_{2}SO_{4}you need twice that many moles of NaOH to neutralize all the H_{2}SO_{4}.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 / concentrationV = 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:**

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 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 H_{3}Cit for the formula.H

_{3}Cit + 3NaOH ---> Na_{3}Cit + 3H_{2}OThe key is the 1 : 3 molar ratio between H

_{3}Cit and NaOHmoles 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 H

_{3}Cit)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 H_{2}SO_{4} solution. What is the concentration of H_{2}SO_{4}?

**Solution:**

2Al(OH)_{3}+ 3H_{2}SO_{4}---> Al_{2}(SO_{4})_{3}+ 6H_{2}OAl(OH)

_{3}and H_{2}SO_{4}are in a 2 : 3 molar ratio.moles Al(OH)

_{3}---> (0.250 mol/L) (20.00 mL) = 5.00 millimoles2 is to 3 as 5.00 is to x

x = 7.50 millimoles

molarity H

_{2}SO_{4}---> 7.50 millimoles / 75.00 mL = 0.100 M

**Problem #10:** 51.00 ml of phosphoric acid solution (H_{3}PO_{4}) 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:**

2H_{3}PO_{4}+ 3Ba(OH)_{2}---> Ba_{3}(PO_{4})_{2}+ 6H_{2}Omoles Ba(OH)

_{2}: 13.90 g / 171.344 g/mol = 0.08112335 mol3 moles Ba(OH)

_{2}react with 2 moles H_{3}PO_{4}0.08112335 mol Ba(OH)

_{2}react with x mol H_{3}PO_{4}x = 0.0540822 mol

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

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