1a) Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in

Question

1a) Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP. How many grams of KHP are needed to exactly neutralize 27.4 mL of a 0.593 M barium hydroxide solution ?

1b) Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP. What volume of a 0.595 M barium hydroxide solution is needed to exactly neutralize 8.33 grams of KHP ?

1c) 57.7 mL of 1.19 M hydrobromic acid is added to 49.5 mL of calcium hydroxide, and the resulting solution is found to be acidic. 19.4 mL of 1.37 M barium hydroxide is required to reach neutrality. What is the molarity of the original calcium hydroxide solution? 1d) 46.8 mL of 0.641 M hydroiodic acid is added to 49.7 mL of potassium hydroxide, and the resulting solution is found to be acidic. 29.1 mL of 0.295 M barium hydroxide is required to reach neutrality. What is the molarity of the original potassium hydroxide solution?

1d) 46.8 mL of 0.641 M hydroiodic acid is added to 49.7 mL of potassium hydroxide, and the resulting solution is found to be acidic.

29.1 mL of 0.295 M barium hydroxide is required to reach neutrality.

What is the molarity of the original potassium hydroxide solution?

Explanation / Answer

Ans. #a. 2 KHC8H4O4(aq) + Ba(OH)2(aq) -------> Ba(KC8H4O4)2(aq) + 2 H2O(l)

According to the stoichiometry of balanced reaction, 1 mol Ba(OH)2 neutralizes 2 mol KHP.

# Moles of Ba(OH)2 to be neutralized = Molarity x Volume of solution in liters

                                                = 0.593 M x 0.0274 L

                                                = 0.0162482 mol

# Required moles of KHP = 2 x Moles of Ba(OH)2 to be neutralized

                                                = 2 x 0.0162482 mol

                                                = 0.0324964 mol

Now,

            Required mass of KHP = Required moles of KHP x Molar mass

                                                = 0.0324964 mol x (204.22 g/ mol)

                                                = 6.6364 g

#b. Moles of KHP taken = 8.33 g / (204.22 g/ mol) = 0.040789 mol

# Following stoichiometry, required moles of Ba(OH)2 = ½ x Moles of KHP

                                                = ½ x 0.040789 mol

                                                = 0.0203945 mol

# Required volume of Ba(OH)2 = Required moles / Molarity

                                                = 0.0203945 mol / (0.595 M)

                                                = 0.0203945 mol / (0.595 mol/ L)

                                                = 0.03428 L

                                                = 34.28 mL

#c. Step 1: Neutralization of excess HBr with Ba(OH)2

Balanced reaction:    HBr(aq) + Ba(OH)2(aq) ----> BaBr2(aq) +2 H2O(l)

According to the stoichiometry of balanced reaction, 1 mol Ba(OH)2 neutralizes 2 mol HBr.

# Moles of Ba(OH)2 consumed = 1.37 M x 0.0194 L = 0.026578 mol

So, moles of excess HBr = 2 x Moles of Ba(OH)2 consumed

                                                = 2 x 0.026578 mol

                                                = 0.053156 mol

# Step 2: Total initial moles of HBr taken = 1.19 M x 0.0577 L = 0.068663 mol

Mol of HBr consumed to neutralize Ca(OH)2 = Total HBr moles – Moles of excess HBr

                                                = 0.068663 mol - 0.053156 mol

                                                = 0.015507 mol

# Step 3: Neutralization of Ca(OH)2 with HBr   

Balanced reaction:    HBr(aq) + Ca(OH)2(aq) ----> CaBr2(aq) +2 H2O(l)

According to the stoichiometry of balanced reaction, 1 mol Ca(OH)2 neutralizes 2 mol HBr.

So,

Moles of Ca(OH)2 in sample = ½ x Moles of HBr consumed by Ca(OH)2 sample

                                                = ½ x 0.015507 mol

                                                = 0.0077535 mol

# Now, molarity of Ca(OH)2 = Moles / Volume of solution in liters

                                                = 0.0077535 mol / 0.0495 L

                                                = 0.1566 M

#d. Step 1: Neutralization of excess HI with Ba(OH)2

Balanced reaction:    HI(aq) + Ba(OH)2(aq) ----> BaI2(aq) +2 H2O(l)

According to the stoichiometry of balanced reaction, 1 mol Ba(OH)2 neutralizes 2 mol HI.

# Moles of Ba(OH)2 consumed = 0.295 M x 0.0291 L = 0.0085845 mol

So, moles of excess HI = 2 x Moles of Ba(OH)2 consumed

                                                = 2 x 0.0085845 mol

                                                = 0.017169 mol

# Step 2: Total initial moles of HI taken = 0.641 M x 0.0468 L = 0.0299988 mol

Moles of HI consumed to neutralize KOH = Total HI moles – Moles of excess HBr

                                                = 0.0299988 mol - 0.017169 mol

                                                = 0.0128298 mol

# Step 3: Neutralization of KOH with HI

Balanced reaction:    HI(aq) + KOH(aq) ----> KI(aq) +H2O(l)

According to the stoichiometry of balanced reaction, 1 mol KOHneutralizes 1 mol HI.

So,

Moles of KOH in sample = 0.0128298 mol = Moles of HI consumed to neutralize it

# Now, molarity of KOH = Moles / Volume of solution in liters

                                                = 0.0128298 mol / 0.0497 L

                                                = 0.2581 M

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