Question

sodium bicarbonate 10 g sodium carbonate 12.72 g, add-50 ml of deionized water to the beaker and stir until all of the weak acid or weak base has dissolved. Carefully and slowly pour the weak acid or weak base solution into the volumetric flask.

preparation of Buffer

pour your weak acid and your weak base together into a bottle.

How many grams of sodium bicarbonate (M.W.= 84.01 g/mole) would you have had to weigh out to make up 500ml of a 0.300M solution of this weak acid?

It takes 0.500liters of 4.00 M Naoh (lye) solution to clean a drain. How many grams of Naoh would have to be weighed out?

How many grams of any solid would you need to weigh out to make 2.5L 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 conjugate base, sodium carbonate.

How many grams of weak acid should be weighed out? How many grams of conjugate base should be weighed out?

Do I put acid first in the water first or base in the water first? Can you write more legibly please?

Answer 1

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 0.300M solution of weak acid the amount of sodium bi carbonate required is

W = 0.3 x 500 x 84.01 / 1000 [ given molecular weight of sodium bi carbonate = 84.01 g/mol]

W = 12.60 grams

Therefore, the require amount of sodium bi carbonate to prepare a 500ml of 0.3M solution of this weak acid is 12.60 grams.

bA)

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 4.0 M solution of NaOH,the amount of sodium hydroxide required is

W = 4.0 x 500 x 40 / 1000 [ molecular weight of sodium hydroxide = 40.0 g/mol]

W = 80 grams

Therefore, the require amount of sodium hydroxide to prepare a 500ml of 4.0M solution of NaOH is 80.0 grams.

cA)

We know that % solution means gm/100ml

4% solution = 4.0 gms/100 ml

for 100ml ---- 4.0 gms

for 2500ml ---- X gms (say)

so X = 2500 x 4.0 / 100 = 100gms.

Therefore, the require amount of solid to prepare a 2.5L(2500ml) of 4.0% solution is 100.0 grams

dA)

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 100ml of 0.2 M solution of weak acid the amount of sodium bi carbonate required is

W = 0.2 x 100 x 84.01 / 1000 [ given molecular weight of sodium bi carbonate = 84.01 g/mol]

W = 1.68 grams

Therefore, the require amount of sodium bicarbonate(weak acid) to prepare a 100ml of 0.20M solution is 1.68 grams

Now, for preparing 100ml of 0.2 M solution of its conjugate base the amount of sodium carbonate required is

W = 0.2 x 100 x 106 / 1000 [ given molecular weight of sodium carbonate = 106 g/mol]

W = 2.12 grams

Therefore, the require amount of sodium carbonate(conjugate base) to prepare a 100ml of 0.20M solution is 2.12 grams