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what mass of barium hydroxide is needed to make 325.0mL of solution that contains a hydroxide ion concentration of 0.425 mol/L ?

Sagot :

[tex]Ba(OH)_2\rightarrow Ba^{2+}\text{ + 2OH}[/tex]

We are required to calculate the mass of Ba(OH)2 needed.

Given:

V = 325.0 mL = 0.350 L

C = 0.425 mol/L

Molarity = n/V (C = n/V)

where n is the moles of solute and V is the volume of solution.

So we need to get the number of moles of OH first then use stoichiometry to find the mass of Ba(OH).

number of moles of OH:

n = Cv

n = 0.425 x 0.350

n = 0.138 mol

The ratio of Ba(OH)2 to OH is 1:2

Therefore number of moles of Ba(OH)2 = 0.138 x (1/2)

number of moles (n) of Ba(OH)2 = 0.069 moles.

Now to get the mass we will use the following equation:

n = m/M where m is the mass and M is the molar mass of Ba(OH)2

We know that the molar mass of Ba(OH)2 = 171.34 g/mol

m = nM

m = 0.069 x 171.34

m = 11.83 g

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