A solution of HClO4 was standardized by dissolving 0.4125 g of primary standard grade HgO ina solution of KBR: HgO (s) + 4Br- + H2O → HgBr4 + 2OH-

The liberated OH- consumed 46.51 mL of the acid. calculate the molarity of the HClO4

0.08190M

mols primary std = g/molar mass = 0.4125/HgO = approximately 0.002 but you need to do that more accurately.

Convert mols HgO to mols OH^- = approx 0.004 (again redo all of this).

HClO4 + OH^- ==> H2O + ClO4^-
0.004 mols OH will consume 0.004 mol HClO4

M HClO4 = mols HClO4/L HClO4. = approx 0.004/0.04651 = ? (approx)

139

Well, let's dive into some chemistry math and humor! To find the molarity of HClO4, we need to calculate the number of moles of OH- that reacted with the acid.

First, we need to convert the volume of the acid solution (46.51 mL) to liters. We divide it by 1000 because jokes about metric conversions are always a Lit-re League of their own!

46.51 mL ÷ 1000 = 0.04651 L

Now, in the balanced equation, we can see that the ratio of OH- to HClO4 is 2:1. So, the number of moles of OH- is half the number of moles of HClO4. If you want to make this equation laugh, just add a "2-bu" soda!

So, to find moles of OH-:

moles of OH- = 0.04651 L ÷ 2 = 0.023255 moles

We know that 0.4125 g of HgO reacts with the same number of moles of OH-. Now, that's some balancing act!

Using the atomic weight of HgO (being 216.59 g/mol) and some hilarious mole-to-mole conversions, we can calculate the number of moles of HClO4:

moles of HClO4 = (0.4125 g HgO ÷ 216.59 g/mol) ÷ 2 = 0.0009532 moles

Finally, we can find the molarity (M) of HClO4 by dividing the moles of HClO4 by the volume (in liters) of the acid solution. Jokes about concentration never get diluted, right?

Molarity of HClO4 = 0.0009532 moles ÷ 0.04651 L ≈ 0.0205 M

So, the molarity of HClO4 is approximately 0.0205 M. Remember, in the world of chemistry, it's all about balancing equations and having a good sense of humor!

To calculate the molarity (M) of HClO4, we need to gather some information and follow a few steps:

1. Calculate the moles of HgO that reacted:
Given the mass of HgO is 0.4125 g and its molar mass is 216.59 g/mol, we can use the formula:
Moles = Mass / Molar mass
Moles of HgO = 0.4125 g / 216.59 g/mol

2. Determine the moles of OH- consumed:
Since the balanced equation shows that 1 mole of HgO produces 2 moles of OH-, the moles of OH- consumed would be:
Moles of OH- = 2 * Moles of HgO

3. Calculate the molarity of HClO4:
The moles of HClO4 present in the solution can be calculated using the stoichiometry of the reaction between OH- and HClO4.
Since 1 mole of HClO4 reacts with 2 moles of OH-, we can use the formula:
Molarity (M) = Moles of HClO4 / Volume of HClO4 (in liters)
The volume of HClO4 is given as 46.51 mL, which is equivalent to 0.04651 L.

Therefore:
Molarity of HClO4 = Moles of HClO4 / Volume of HClO4 (in liters)
= Moles of OH- / Volume of HClO4 (in liters)

Note: The moles of OH- consumed are twice the moles of HClO4 in the reaction, as per the balanced equation.

Plug in the values to calculate the molarity of HClO4.