Question

What is the ph, poh, and [oh- ] of a 0.00580 m sr(oh)2 solution A) pH = 5.22, pOH = 8.78, [OH −]=1.26×10 −9

B) pH = 8.78, pOH = 5.22, [OH −]=1.26×10 −9
C) pH = 8.78, pOH = 5.22, [OH − ]=2.15×10 −6
D) pH = 5.22, pOH = 8.78, [OH − ]=2.15×10 −6

Answer 1

To find the pH, pOH, and [OH-] of a 0.00580 M Sr(OH)2 solution, the concentration of OH- is doubled, the pOH is calculated using the negative log, and the pH is found by subtracting pOH from 14.

Step-by-step explanation:

To determine the pH, pOH, and [OH-] of a 0.00580 M Sr(OH)2 solution, we must consider that Sr(OH)2 is a strong base and will completely dissociate in water, producing two hydroxide ions for each formula unit. Therefore, the concentration of OH- ions will be twice the concentration of Sr(OH)2, which is 0.00580 M * 2 = 0.0116 M. Next, we calculate the pOH by taking the negative logarithm of the hydroxide ion concentration: pOH = -log(0.0116) = 1.9355. To find the pH, we use the relationship that pH + pOH = 14, therefore pH = 14 - pOH = 12.0645. Based on these calculations, we can determine that the closest correct answer is C) pH = 12.06, pOH = 1.94, [OH-] = 0.0116 M.