Question

What is the molarity of 555 l of a ba(oh)2 solution if the ph is 10.20? 1. 2.26 × 10−5 m 2. 6.31 × 10−11 m 3. 5.15 × 10−7 m 4. 3.15 × 10−11 m 5. 3.14 × 10−4 m 6. 1.58 × 10−4 m 7. 7.92 × 10−5 m 8. 4.40 × 10−2 m?

Answer 1

7. 7.92 × 10−5

Step-by-step explanation:

Hello,

In this case, with the given pH, one could find the pOH:

`pOH=14-pH=14-10.20=3.8`

Thus, since barium hydroxide is completely dissolved in water based on:

`Ba(OH)_2\rightarrow Ba^(+2)+2OH^-`

The concentration of hydroxyl ions is twice to that of the hydroxide (2:1 mole relationship). Therefore, by considering the relationship between the pOH and the concentration of hydroxyl we have:

`pOH=-log([OH]^-)\\`

`[OH]^-=10^(-3.8)=1.58x10^(-4)M`

Finally, given the 1:2 mole ratio of barium hydroxide to hydroxyl ions, the concentration of barium hydroxide results:

`[Ba(OH)_2]=2*[OH^-]=(1)/(2) *1.58x10^(-4)M`

`[Ba(OH)_2]=7.92x10^(-5)M`

Thus, the answer is 7. 7.92 × 10−5.

Regards.

Answer 2

First, you need to get POH from the value of PH:

when POH = 14 - PH

= 14 - 10.2

= 3.8
then we are going to get the value of [OH] from the POH value:

POH = -㏒[OH-]

3.8 = - ㏒ [OH-]

∴[OH-] = 1.58 x 10^-4

then, we will get the moles of ba(OH)2 = (1.58 x 10^-4) / 2

= 0.0000792 moles

∴ the molarity of Ba(OH)2 = 7.92 x 10^-5