Describe how you would prepare the following in aqueous solution. a) 0.50M solut

Question

Describe how you would prepare the following in aqueous solution.

a) 0.50M solution of HCl that would just neutralize 5.0g of Ba(OH)2 starting with 6.0 M HCl

This is a 2 part question in volume. The first part is V6.0M HCl (in ml) and the second is V0.5 M HCl. Thank you

******Please Im struggleing here Id greatly appreciate some indepth explaination not just giving the anwer. Id like to know how to do it for the test too! THANK YOU THANK YOU!!!   ;P

PLEASE DONT USE THE ANSWERS FROM YAHOO THAT ARE FROM 5 years AGO... ive already tried it... I NEED real help....

Soln Hcl, nevet 5.9 Ba(OH)2, start w/6.M Hcl mass of Ba(OH2) 5.0g Mol " " = 0.029181471 g/mol mol of Hcl needed x2 = 0.058362942 mol Hcl/5L = 0.116725884L Then tl used M1V1 = M2V2

Explanation / Answer

continued from previous post

mol of HCl needed = 2 * 0.0291889 = 0.058378 mol

as per the calculations shown in previous post

Volume of 6.0 M HCl needed = 9.73 mL

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Now for volume of 0.50 M HCl

after dilution of 6.0 M HCl you get 107.30 mL of 0.50 M HCl solution

Now what volume of this solution is need to neutralize 5.0 g Ba(OH)2 ?

mol of HCl needed = 2 * 0.0291889 = 0.058378 mol

volume of 0.50 M solution containing 1 mole HCl = 1000/0.50 mL

volume of 0.50 M HCl solution containing 0.058378 mol HCl = 0.058378 * 1000/0.50 = 1167.56 mL

Notice that in question you mentioned 0.5 M however in coment you have mentioned 5.0 M.

Thus for 5.0 M HCl here is the answer

volume of 5.0 M HCl solution containing 0.058378 mol HCl = 0.058378 * 1000/5.0 = 116.76 mL

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