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**Problem #1:** Consider the titration of 80.0 mL of 0.100 M Ba(OH)_{2} by 0.400 M HCl? What is the pH of the solution (a) before adding any acid and (b) after adding 20.0 mL of HCl?

**Solution:**

1) Determine [OH¯], then pH:

Ba(OH)_{2}---> Ba^{2+}+ 2OH¯Ba(OH)

_{2}is a strong base, so it dissociates 100%. Remember also, that two hydroxides are produced for every Ba(OH)_{2}.[OH¯] = 0.200 M

pOH = -log 0.200 = 0.699

pH = 14.000 - pOH = 14.000 - 0.699 = 13.301 (the answer to part a)

2) Determine moles of OH¯ remaining after 20.0 mL of HCl is added:

moles H^{+}⇒ 0.400 mol/L x 0.0200 L = 0.00800 mol(Remember, HCl dissociates 100%.)

moles [OH¯] ⇒ 0.200 mol/L x 0.0800 L = 0.0160 mol

(Note use of 0.200 M for [OH¯]; the [Ba(OH)

_{2}] is 0.100 M.)remaining OH¯ ⇒ 0.0160 mol - 0.00800 mol = 0.00800 mol

3) Determine [OH¯], then pH:

[OH¯] = 0.00800 mol / 0.100 L = 0.080 M(Note the use of 100 mL, which comes from 80 + 20. There is an asumption that the volumes are additive. In context - both solutions are aqueous - this is a pretty safe assumpton to make.)

pOH = -log 0.080 = 1.097

pH = 14.000 - 1.097 = 12.903 (the answer to part b)

**Problem #2:** 32.00 g of sodium hydroxide were dissolved in 250.0 mL of solution to prepare the titrant. 25.00 mL of sulfuric acid were titrated with above titrant. It took 16.00 mL of sodium hydroxide solution to titrate to the end point. What is the molarity of the sulfuric acid?

**Solution:**

1) Determine molarity of NaOH:

MV = grams divided by molar mass(x) (0.2500 L) = 32.00 g / 40.00 g/mol

x = 3.200 M

2) Determine moles NaOH in 16.00 mL of 3.200 M solution:

moles NaOH ⇒ (3.200 mol/L) (0.01600 L) = 0.05120 mol

3) Determine moles of H_{2}SO_{4} that react:

2NaOH + H_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}OThe molar ratio between NaOH and H

_{2}SO_{4}is 2:1Therefore, 0.05120 moles of NaOH will neutralize 0.02560 moles of H

_{2}SO_{4}

4) Calculate molarity of H_{2}SO_{4} solution:

0.02560 mol / 0.02500 L = 1.024 M

**Problem #3:** Consider the reaction:

2HCl + Ba(OH)_{2}---> BaCl_{2}+ 2H_{2}O

How many mL of 0.7000 M HCl solution would just react with 8.000 g of Ba(OH)_{2}?

**Solution:**

1) Calculate moles of Ba(OH)_{2}:

8.000 g / 171.3438 g/mol = 0.04669 mol

2) Determine moles of HCl required to neutralize:

The molar ratio between HCl and Ba(OH)_{2}is 2:1Therefore, 0.09338 moles of HCl is required to neutralize 0.04669 moles of Ba(OH)

_{2}

3) Determine volume of HCl required:

0.7000 mol/L = 0.09338 mol / x x = 0.1334 L = 133.4 mL

**Problem #4:** A 50.0 mL sample of 0.50 M HCl is titrated with 0.50 M NaOH. What is the pH of the solution after 28.0 mL of NaOH have been added to the acid?

**Solution:**

1) Calculate moles of HCl and NaOH:

moles HCl ⇒ (0.50 mol/L) (0.050 L) = 0.025 mol

moles NaOH ⇒ (0.50 mol/L) (0.028 L) = 0.014 mol

2) Calculate moles HCl remaining:

Since HCl and NaOH react in a 1:1 molar ratio:0.025 mol - 0.014 mol = 0.011 mol HCl remaining

3) Calculate [HCl] of new solution:

0.011 mol / 0.078 L = 0.141 MNote volume of 78 mL, derived from 50 + 28.

4) Calculate pH:

pH = -log [H^{+}]Since HCl dissociates 100%:

pH = -log 0.141 = 0.85

**Problem #5:** How much anhydrous CaO would be needed to neutralize 0.900 L of 5.00 M H_{2}SO_{4}?

**Solution:**

1) CaO reacts with water as follows:

CaO + H_{2}O ---> Ca(OH)_{2}

2) Calcium hydroxide racts with sulfuric acid as folows:

Ca(OH)_{2}+ H_{2}SO_{4}---> CaSO_{4}+ 2H_{2}O

3) Combining them gives:

CaO + H_{2}SO_{4}---> CaSO_{4}+ H_{2}OThe important point is that there is a 1:1 molar ratio between CaO and H

_{2}SO_{4}

4) Determine moles of H_{2}SO_{4}:

(0.900 L) (5.00 mol/L) = 4.50 mol

5) Determine grams of CaO required:

4.50 mol times 56.077 g/mol = 252 g (to three sig fig)

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