ksp values and the Ksp dissociation equilibrium:View attachment 39174
Answer is D
How would you solve this question using calculations (of course just by looking at it it's D cause basic pH, but im curious about calc)
Thanks
That's my best guess at least, although I kinda tried it now and couldn't get any of the given answersksp values and the Ksp dissociation equilibrium:
You are given that
View attachment 39175 on the data sheet
Therefore [Ca2+][OH]^2 = 5.02*10^-6
Therefore given Ca concentration is 0.0015, you can solve for OH-
hmmmm I still assume it still has something to do with KspOh ok cheers, though doing it that way gives 12.76 and assumes Ca2+ conc is same as Ca(OH)2
I tried using RICE table but got relatively close to answer but not it
I feel sillyu know that Ca(OH)2 <--> Ca2+ + 2OH-
so pH = -log(0.0015 x 2)
pOH = 14 - that value
u get 11.48
u dont need to use Ksp
its okay ur just overthinking itI feel silly
Oh ok thanks, but can you say that the conc of hydroxide ions is in molar ratio to Ca(OH)2 though (since 0.0015 is initial conc., would you need to logically do RICE)u know that Ca(OH)2 <--> Ca2+ + 2OH-
so pH = -log(0.0015 x 2)
pOH = 14 - that value
u get 11.48
u dont need to use Ksp
no u don't even need to do rice, u just need to know that there's 2 of OH so u just multiply the given conc by 2Oh ok thanks, but can you say that the conc of hydroxide ions is in molar ratio to Ca(OH)2 though (since 0.0015 is initial conc., would you need to logically do RICE)
Oh ok thanksno u don't even need to do rice, u just need to know that there's 2 of OH so u just multiply the given conc by 2