Answer:
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)
Volume of 0.85M Ba(OH)₂ needed = 4.45 ml and dilute to 3.00 Liters
pH is a measure of how acidic/basic water is. The range goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.
pH is really a measure of the relative amount of free hydrogen and hydroxyl ions in the water
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)
Hence Volume of 0.85M Ba(OH)₂ needed = 4.45 ml and dilute to 3.00 Liters
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