Question

What are volumetric pipets and burets and what are they used for? 1. 2. To what volume are you to record the values from the buret? 3. 0.3811 g sample of KOH will just neutralize what volume of 0.2000 M HS02 Write the balanced chemical equation. a. b. What volume of 0.2000 M H SO, will just neutralize the KOH sample? 4. In a experiment similar to yours, a solution is prepared by adding 4.512 g of slaked lime also known as calcium hydroxide to 500.00 mL of water. What is the molarity of the solution? a. What is the molarity of the ions in the solution? b. c. Write the balanced chemical reaction between calcium hydroxide an You are given a solution of perchloric acid of unknown concen d. calcium hydroxide solution to completely neutralize 15.00 ml. or what is the molarity of the perchloric acid? tely e. The actual molarity of the perchloric acid 0.4237 M, what is the percent error?

Answer 1

1. Volumetric pipets and burets are graduated glassware used in titrimetric analyses. A volumetric buret has a long glass tube with a stopcock at the lower end and a tapered capillary tube following the stopcock. The flow of the liquid from the buret is controlled by the stopcock.

A volumetric pipette on the other hand, has a large bulb at the middle and narrow tubes above and below the bulb. The pipette shows a single graduation mark and can deliver precise volumes of liquids during a titration.

Both the volumetric buret and the pipette are used to transfer accurate volume of liquids. Both the glassware show high levels of accuracy and are invariably used in volumetric titrations.

2. A volumetric buret has the lowest graduation of 0.1 mL. Hence, the volume of a liquid delivered by a volumetric buret can be read to 0.1 mL of a particular value.

3a. The balanced chemical equation for the reaction between KOH and H₂SO₄ is given as

2 KOH (aq) + H₂SO₄ (aq) --------> K₂SO₄ (aq) + 2 H₂O (l)

b. As per the stoichiometric equation,

2 mols KOH = 1 mol H₂SO₄.

The atomic masses are

K: 39.098 u

O: 15.999 u

H: 1.008 u

The gram molar mass of KOH = (1*39.098 + 1*15.999 + 1*1.008) g/mol

= 56.105 g/mol.

Mol(s) KOH corresponding to 0.3811 g = (0.3811 g)/(56.105 g/mol)

= 0.0067926 mol.

As per the balanced equation,

mols H₂SO₄ neutralized = (0.0067926 mol KOH)*(1 mol H₂SO₄)/(2 mols KOH)

= 0.0033963 mol.

The molarity of H_2SO₄ = 0.2000 M.

Therefore, the volume of H₂SO₄ required = (0.0033963 mol)/(0.2000 M)

= 0.0169815 L

= (0.0169815 L)*(1000 mL)/(1 L)

= 16.9815 mL

≈ 17.00 mL (ans).