Question 17

[GraemeMatt8033]

logic for this one is sound. It's just saying that because the dissociation constants of the last two protons are so small, that such a small amount of hydrogen dissociates that the amount it adds to solution is negligible and won't contribute to the overall ph of the solution

[wesley.pau5578]

How is the sq rt of (7.5 x 10^(-3)) approximately 10^(2.2)? Or, how do you get that?

I approximated the sq rt by changing (7.5 x 10^(-3)) to (75 x 10^(-4)), and I approximated the sq rt of that, which I said was: (8.5 x 10^(-2)). However, that didn't seem to work. So how does your approximation work?

[jellywing_2058]

This is how I did it and it works!

Ka1 = [H+][H2PO4-]/[H3PO4]

We know that [H+]=[H2PO4]= x
and that [H3PO4] = 1 M

Ka1 = [H+][H2PO4-]/[H3PO4]
7.5 x 10^(-3) = x^2/1
7.5 x 10^(-3) = x^2
[7.5 x 10^(-3)]^(1/2) = x

Since we do not have a calculator, I approximated the value 7.5 to 10 as such:

[7.5 x 10^(-3)]^(1/2) = x
[10 x 10^(-3)]^(1/2) = x
[10^1 x 10^(-3)]^(1/2) = x
[10^(-2)]^(1/2) = x
10^-1 = x = [H+]

pH = -log [H+]
pH = -log (10^-1)
pH = (-1)(-log 10)
pH = log 10
pH ~ 1

So the answer choice closest to pH 1 is A.

[btdumford5156]

How is the square root of 10^-2.2 found?