What volume of 0.850 M barium hydroxide solution do you need to make 3.00 L of a solution with a pH of 11.4002

Volume of 0.85M Ba(OH)₂ needed = 4.45 ml and dilute to 3.00 Liters

Explanation:

?Vol of 0.85M Ba(OH)₂ => 3.00 Liters solution with pH = 11.4002?

pH = 11.4002 => pOH = 14 - 11.4002 = 2.5998

[OH⁻] = 10⁻²°⁵⁹⁹⁸ M = 2.513 x 10⁻³ M

Since [OH⁻] is 2 times concentration of Ba(OH)₂ then ...

[Ba(OH)₂] = 1/2[OH⁻] = 1/2(2.513 x 10⁻³)M = 1.257 x 10⁻³ M

Using Dilution Equation ...

Molarity(concentrate) x Volume(concentrate) = Molarity(dilute) x Volume (dilute)

=> Vol(dilute needed) = M(c) x V(c)/M(d) = (1.251 x 10⁻³M)(3000ml)/(0.85M)

= 4.45 ml needed

Mixing => 4.45 ml of 0.85M Ba(OH)₂ + water solvent up to but not to exceed 3000ml.

Test Calculation:

Given 4.45ml of 0.85M Ba(OH)₂ solution dilute to 3000ml and calculate pH.

[Ba(OH)₂] = moles Ba(OH)₂/Vol Soln in Liters = 0.00445L x 0.85M /3.00L = 0.0012608M => [OH⁻] = 2 x 0.0012608M = 0.002523M in OH⁻

pOH = -log[OH⁻] = -log(0.002523) = 2.598

pH = 14 - 2.598 = 11.4017 (close enough)

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