A student is asked to prepare 750mL of an aqueous 0.18M NaOH solution. The stude

Question

A student is asked to prepare 750mL of an aqueous 0.18M NaOH solution. The student is provided with water, solid NaOH, a scale, a graduated cylinder and a container which holds exactly 750mL. Explain the steps required (including calculations!) to prepare this solution. 3. If the student was asked to prepare this same solution (750mL of 0.18M NaOH) using a concentrated 6.0M NaOH solution instead of solid NaOH and water, explain the steps (including calculations) the student would take to make the solution. a.

Explanation / Answer

Q3

M= mol/V

mol = M*V = 0.18*0.75 = 0.135 mol of NAOH

mass = mol*MW = 0.135*40 = 5.4 g of NaOH

then...

measure 5.4 g of NaOH in the balance,

Then, add about 200 mL in the 750 mL container...

then add 5.4 g slowly, since the reaction of NaOH in water is exothermic, will increase in temperature

then

after all mass is dissolved, add all other volume up to V= 750 mL

this is now [NaOH] = 0.18 M

a)

if it is C1 = 6M fo NaOH (stock solution)

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

Cstock *Vstock = Cnew * Vnew

6*VStock = 750*0.18

Vstok = 750*0.18/6

Vstock = 22.5 mL

then...

Measure 22.5 mL with a pippette

measure V = 700 mL of water

add it to the solution, dropwise,

eventually, it will be approx 700+22.5 mL = 722.5 mL (soluitoin)

add wate rup to V = 750 mL, this will be 0.18 M in concnetration

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