The pH of the solution is 11.6
given that :
molarity of Ba(OH)₂ = 0.020 M
volume = 50 mL = 0.05 L
number of moles = molarity × volume in L
= 0.020 × 0.05
= 0.001 mol
the molarity after dilution = 0.001 / 0.250
= 0.004
pOH = -log (0.004 )
= 2.39
pH + pOH = 14
pH = 14 - pOH
= 14 - 2.39
= 11.6
Thus, the pH of a solution prepared by diluting 50.00 ml of 0.020 m Ba(OH)₂ with enough water to produce a total volume of 250.00 ml is 11.6.
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