100. mL of 0.100 M HCl aqueous solution is mixed with 200. mL of 0.100 M CaCl2 solution to make a 300.mL solution. Find the molarity of the Cl- and pOH in the mixed solution Question 7 options:

Explanation:

HCl    ⇄    H⁺    +   Cl⁻

HCl is a strong electrolyte so it will ionize completely .

moles of Cl⁻ in 100 mL of .1 HCl = .1 x .1 = .01 gram - ion .

CaCl₂ = Ca⁺² + 2 Cl⁻

1 mole                2 mole

moles of CaCl₂ in 200 mL of .1 M CaCl₂ = .2 x .1 = .02 gram mole  .

.02 gram mole of CaCl₂ will give 2 x .02 = .04 gram-ion of Cl⁻ ion .

Total gram - ion of Cl⁻ = .01 + .04 = .05 gram-ion .

Total volume = 300mL = .3 L

molarity of Cl⁻ = .05 / .3 = .167 M .

HCl    ⇄    H⁺    +   Cl⁻

moles of H⁺ = moles of Cl⁻ = .01 gram-ion .

volume of solution = .3 L

molarity of H⁺ = .01 / .3 = 33.33 x 10⁻³ M .

[ H⁺] [ OH⁻] = 10⁻¹⁴

33.33 x 10⁻³  [ OH⁻] = 10⁻¹⁴

[ OH⁻] = 3 x 10⁻¹³

pOH = - log[OH⁻] = - log3 x 10⁻¹³

13-log3

= 13 - .477

12.52

or ,

pH = - log[H⁺] = - log33.33 x 10⁻³

3 - log 33.33

= 3 - 1.52

= 1.48

pOH = 14 - 1.48 = 12.52

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun

a is the answer to this question

explanation: