I totally forgot how to calculate molarity and all my other resources are not helping.

Molarity is mol of solute/L of solution
So I have 0.1 M of sodium thiosulfate and 0.1 M of hydrochloric acid

For a lab I put in 5 drops of sodium thiosulfate and 3 drops of hydrochloric acid.

I need to calculate the molarity of sodium thiosulfate with the equation M1V1=M2V2 (I believe the amount of drops is volume)

Would it look like this M1*8 = 0.8*8
As that doesn't look correct
Any help would be appreciated :)

M1V1 = M2V2

M 1 = 0.1 V1 = 5 drops
M2 = ? V2 = 8 drops (that's 3 drops + 5 drops = 8 drops)
0.1 x 5 = M2 x 8
That M1V1 = M2V2 comes in handy in many problems but for simple dilution problems like this I prefer to think it out; i.e.,
You have a solution that is 0.1 M and you dilute it with 5 drops up to a total of 8 drops. So final concentration is 0.1 x 5/8 = xxx
For the HCl the final concn is 0.1 x 3/8 = ?

Thank you! Really appreciate it!

To calculate the molarity of sodium thiosulfate (M1), you can use the equation M1V1 = M2V2, where V1 is the volume of sodium thiosulfate you added (in liters), M2 is the molarity of hydrochloric acid, and V2 is the volume of hydrochloric acid you added (in liters).

First, you need to convert the number of drops to liters. You can use the conversion factor: 20 drops = 1 mL.

Given that you added 5 drops of sodium thiosulfate, the volume (V1) can be calculated as follows:
V1 = 5 drops * (1 mL / 20 drops) * (1 L / 1000 mL)
V1 = 0.25 mL * (1 L / 1000 mL)
V1 = 0.00025 L

Similarly, for the hydrochloric acid, if you added 3 drops:
V2 = 3 drops * (1 mL / 20 drops) * (1 L / 1000 mL)
V2 = 0.15 mL * (1 L / 1000 mL)
V2 = 0.00015 L

Now, we can calculate the molarity of sodium thiosulfate:
M1 * 0.00025 L = 0.1 M * 0.00015 L

Rearranging the equation to solve for M1:
M1 = (0.1 M * 0.00015 L) / 0.00025 L
M1 = 0.06 M

Therefore, the molarity of sodium thiosulfate is 0.06 M based on the volumes of drops you added.

To calculate the molarity of a solution, you are correct that you can use the formula M1V1 = M2V2. However, in this case, the number of drops is not a volume measurement that can be directly used in the equation. Drops can vary in size, so it is not an accurate measurement for concentration calculations.

To calculate the molarity accurately, it is better to use a more precise volume measurement, such as milliliters (mL) or liters (L). If you have the precise volume measurements for the drops, you can convert them to milliliters or liters. Then, you can proceed with the calculation.

For example, let's assume that each drop is approximately 0.05 mL (you can adjust this value depending on the size of your drops).

Given:
M1 (initial molarity of sodium thiosulfate) = 0.1 M
V1 (volume of sodium thiosulfate) = 5 drops * 0.05 mL/drop = 0.25 mL = 0.00025 L
M2 (final molarity after the reaction) = ?
V2 (final volume of the solution after the reaction) = 3 drops * 0.05 mL/drop = 0.15 mL = 0.00015 L

Now, you can plug these values into the formula M1V1 = M2V2:

(0.1 M) * (0.00025 L) = M2 * (0.00015 L)

Solving for M2, we get:
M2 ≈ (0.1 M) * (0.00025 L) / (0.00015 L)
M2 ≈ 0.1667 M

So, the molarity of sodium thiosulfate after the reaction is approximately 0.1667 M.

Remember, this calculation assumes that the reaction between sodium thiosulfate and hydrochloric acid goes to completion and that the total volume of the solution does not change significantly. Additionally, always double-check your units to make sure they are consistent throughout the calculation.

Note: It's worth mentioning that the number of drops can be highly subjective and imprecise, so it is generally better to work with more precise measurements when calculating molarity.