finding initial molarity hw 12.25

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D-nice1D
Posts: 19
Joined: Fri Sep 28, 2018 12:21 am

finding initial molarity hw 12.25

Postby D-nice1D » Thu Jul 25, 2019 12:42 am

Calculate the initial molarity of Ba(OH)2 and the molarities of Ba2!, OH", and H3O! in an aqueous solution that contains 0.43 g of Ba(OH)2 in 0.100 L of solution.

How would I go about setting this problem up??
Is there an icebox involved?

somyapanchal1D
Posts: 34
Joined: Mon Jun 17, 2019 7:23 am

Re: finding initial molarity hw 12.25

Postby somyapanchal1D » Thu Jul 25, 2019 6:05 pm

No ICE tables will be used since we never went over it in class.
First, you would write out the chemical reaction of Ba(OH)2 and balance it in order to see the moles of each product and reactant. Next, since we are trying to find the molarity, you need to convert grams of Ba(OH)2 to moles. Next, you would use the equation Molarity=moles/liters to find the molarity of the Ba(OH)2 solution. In order to find the molarity of Ba2+ and OH-, you would refer back to the balanced chemical equation. Since one mole of Ba(OH)2 produces one mole of Ba2+, the molarity would be the same as Ba(OH)2. Also, since one mole of Ba(OH)2 produces two moles of OH-, the molarity of OH- would be twice that of Ba(OH)2. Finally, to find the molarity of H3O+, you need to use the formula Kw=[OH-][H3O+] and solve for [H3O+] where Kw is 1*10^-14. Here is my work:
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Kevin To 1B
Posts: 32
Joined: Sat Feb 02, 2019 12:15 am

Re: finding initial molarity hw 12.25

Postby Kevin To 1B » Thu Jul 25, 2019 6:06 pm

First, let's realize that Ba(OH)2 dissociates into Ba2+ and (OH)-. As a chemical equation, we can write:
Ba(OH)2(s) → Ba2+(aq) + 2(OH)-(aq)

(a) Initial Molarity of Ba(OH)2
We just use stoichiometry to calculate the initial molarity. We are given 0.43 g Ba(OH)2 in 0.100 L of solution. Convert the grams of Ba(OH)2 to moles using the molar mass of Ba(OH)2. Then, since Molarity is moles of solute divided by volume of solution, divide the calulated moles by given volume in Liters.

(b) Molarity of Ba2+
Looking at the dissociation formula, we can see that the moles of Ba(OH)2 and Ba2+ are a 1:1 ratio. Therefore, the Molarity of Ba(OH)2 is equal to Ba2+.

(c) Molarity of (OH)-
Same step as previous, but since the ratio of Ba(OH)2 and (OH)- is 1:2, use this as the converting factor. In other words, multiply the molarity Ba(OH)2 by 2 to get the Molarity of (OH)-

(d) Molarity of H3O+
First, know that KW = [H3O+][(OH)-] = 1.0 x 10-14. To find the Molarity of H3O+, substitute the molarity of (OH)- to the KW formula.


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