1. When the hydrochloric acid is added to the calcium carbonate, will there be an excess of
HCl(aq) or CaCO3(aq) or will there be stoichiometric equivalents of the two? Show your
calculations.
2. In this lab, you will use a 50mL buret and ~0.3M NaOH (aq). Calculate the mass of
potassium hydrogen phthalate, KHP, needed for the standardization of NaOH (aq) in Part B.
3. Which indicator would you choose to perform titration in lab and why? What color change do
you expect to see?
Relevant: document. li / 2ZBS
1. Who knows? You must know the mols HCl and mols CaCO3 before you can answer this question.
2. What is part B?
3. Strong base and weak acid, I would use phenolphthalein.
1. To determine if there will be an excess of HCl(aq) or CaCO3(aq) or stoichiometric equivalents of the two when hydrochloric acid is added to calcium carbonate, you can use stoichiometry.
The balanced equation for the reaction is:
CaCO3(aq) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)
From the balanced equation, you can see that 1 mole of CaCO3 reacts with 2 moles of HCl.
To calculate the stoichiometric equivalents, you need to compare the number of moles of HCl with the number of moles of CaCO3. You can do this by using the given quantities of both compounds. Let's say you have 10 mL of 2M HCl(aq) and 10 g of CaCO3.
First, calculate the number of moles of HCl using the concentration and volume:
moles of HCl = concentration x volume
= (2 mol/L) x (0.01 L)
= 0.02 moles
Next, calculate the number of moles of CaCO3 using the mass and molar mass:
moles of CaCO3 = mass / molar mass
= 10 g / (40.08 g/mol + 12.01 g/mol + 3 x 16.00 g/mol)
= 10 g / 100.08 g/mol
= 0.1 moles
Since the reaction requires 2 moles of HCl for every 1 mole of CaCO3, we can see that there is an excess of HCl. This means that some of the HCl will be left over after the reaction is complete.
2. To calculate the mass of potassium hydrogen phthalate (KHP) needed for the standardization of NaOH(aq), you need to consider the balanced equation between the two compounds. Let's assume you want to react 1 mole of NaOH with KHP.
The balanced equation for the reaction is:
NaOH(aq) + C8H5O4K(aq) -> H2O(l) + NaKC8H4O4(aq)
From the equation, you can see that 1 mole of NaOH reacts with 1 mole of KHP.
The molarity of NaOH is given as ~0.3M, which means there are approximately 0.3 moles of NaOH per liter of solution.
To find the mass of KHP needed, you can use the molar mass of KHP. The molar mass of KHP is 204.22 g/mol.
mass of KHP = moles of NaOH x molar mass of KHP
= (0.3 mol/L) x (204.22 g/mol)
= 61.266 g
Therefore, you need approximately 61.266 grams of KHP for the standardization of NaOH(aq) in Part B.
3. The choice of indicator for the titration in the lab depends on the pH range where the equivalence point occurs. One commonly used indicator is phenolphthalein, which turns pink in basic solutions with a pH range of approximately 8.2 to 10.0. This indicator is often used in acid-base titrations because its color change is easily visible.
However, without more specific information about the nature of the titration and the expected pH at the equivalence point, it is difficult to determine the most appropriate indicator. The relevant document you mentioned (document.li/2ZBS) could potentially provide more details and guide you in selecting the appropriate indicator. You should refer to that document to find the recommended indicator and the expected color change at the equivalence point.