ASSIGNMENT. PRACTICE CALCULATIONS 1. Convert units (show your work!): a. 0.01L= ml b. 0.05M= mM c. 0.2M= mM d. 10mM= M 2. Molar mass of NaOH is 40g/moo. How many grams of NaOH do you have to take to prepare: a. 1L of 10M solution of NaOH ? b. 1L of 10mM solution of NaOH ? c. 100ml of 10mM solution of NaOH ? 3. Calculate the theoretical pH of 10mMNaOH using the data below: - NaOH is a strong base, dissociates completely, thus [OH] is equal to [NaOH] - ion product of water [H+][OH]=1.01014M2 - pH=log[H+] 4. Calculate the amount of concentrated (37%)HCl in ml necessary to make 100ml of 10mM solution. HCl is a liquid with molecular weight of 36.5g/mol. Usually HCl is sold as a solution, and figuring out how to prepare necessary dilutions could be tricky. You can proceed to calculations if you already know how to do it; if not, we will try to do it step by step: a. Calculate the theoretical dry mass in grams of HCl to make 100ml of 10mM solution of HCl, as if it was not liquid, but solid. b. Since HCl is sold as liquid, we need to measure the amount of stock HCl in ml, not in grams. If the density of HCl solution is 1.27g/ml, how many ml of liquid 100%HCl do you need to prepare 100ml of 10mM solution of HCl ? Here you will need to take the amount of grams from the previous step and calculate the amount of ml using the provided density. c. Now, you have to adjust your calculations, since you actually have a 37%HCl solution. This is how HCl is normally sold. You will need to increase the amount of ml you calculated in the previous step accordingly to prepare the solution you need. d. Calculate the theoretical pH of this solution ( HCl is a strong acid and dissociates completely). Note: In the lab we normally need to check the jars with sodium salts solids for the molecular weightl You need to use the molecular weight of the salts from the jars! The salts are often sold as hydrates, and you need to use the molecular weight from the jar. We have different jars in the lab, make sure you use salts from the same jar you made the calculations for