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)