sodium bicarbonate 10 g sodium carbonate 12.72 g, add-50 ml of deionized water to the beaker...

90.2K

Verified Solution

Question

Chemistry

sodium bicarbonate 10 g sodium carbonate 12.72 g, add-50 ml ofdeionized water to the beaker and stir until all of the weak acidor weak base has dissolved. Carefully and slowly pour the weak acidor weak base solution into the volumetric flask.
preparation of Buffer
pour your weak acid and your weak base together into abottle.
How many grams of sodium bicarbonate (M.W.= 84.01 g/mole) wouldyou have had to weigh out to make up 500ml of a 0.300M solution ofthis weak acid?
It takes 0.500liters of 4.00 M Naoh (lye) solution to clean adrain. How many grams of Naoh would have to be weighed out?
How many grams of any solid would you need to weigh out to make2.5L of a 4% solution? (Recall, a 4% solution means 4 g/100 mL)
You will make up 100ml of a 0.20 M solution of a weak acid,sodium bicarbonate and 100ml of a 0.20 M solution of its conjugatebase, sodium carbonate.
How many grams of weak acid should be weighed out? How manygrams of conjugate base should be weighed out?
Do I put acid first in the water first or base in the waterfirst? Can you write more legibly please?

Answer & Explanation Solved by verified expert

3.5 Ratings (411 Votes)

aA We know that molarity M weight gram molecular weight x 1000 V in ml so Weight W M x V x gram molecular weight 1000 Now for preparing 500ml of 0300M solution of weak acid the amount of sodium bi carbonate required is W 03 x 500 x 8401 1000 given molecular weight of sodium bi carbonate 8401 gmol W 1260 grams Therefore the require amount of sodium See Answer

Get Answers to Unlimited Questions

Join us to gain access to millions of questions and expert answers. Enjoy exclusive benefits tailored just for you!

Membership Benefits:

Unlimited Question Access with detailed Answers
Zin AI - 3 Million Words
10 Dall-E 3 Images
20 Plot Generations
Conversation with Dialogue Memory
No Ads, Ever!
Access to Our Best AI Platform: Flex AI - Your personal assistant for all your inquiries!

Become a Member